[ { "type": "user", "uuid": "bd1fb947-6d41-46eb-bcd4-a3671dd4dfa0", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "role": "user", "content": [ { "type": "text", "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nSolve from the supplied statement first. For very hard problems, do a quick source check for the official Codeforces tutorial/editorial and accepted submissions when they would help you derive or verify the solution. If a source is blocked, missing, Cloudflare-challenged, 404, or otherwise unavailable, keep solving from the supplied statement and remember that source status. If you still can't solve after that, it's ok.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2247A: \"Zero Sum\".\n\nKnown problem metadata:\n- Contest ID: 2247\n- Index: A\n- Name: Zero Sum\n- Rating: unknown / unrated\n- Tags: constructive algorithms, dp, number theory\n- Problem URL: https://codeforces.com/contest/2247/problem/A\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n\n
A. Zero Sum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given an array $$$a$$$ of length $$$n$$$, consisting only of $$$-1$$$ and $$$1$$$.

You may perform the following operation on $$$a$$$ any number of times:

  1. Choose an index $$$i$$$ satisfying $$$1 \\le i \\le n - 1$$$.
  2. Assign $$$a_i = -a_i$$$ and $$$a_{i + 1} = -a_{i + 1}$$$.

Determine whether it is possible to make the sum of elements of $$$a$$$ equal to $$$0$$$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 100$$$) — the length of array $$$a$$$.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$a_i \\in \\{-1, 1\\}$$$) — the array $$$a$$$.

Output

For each test case, print \"YES\" if it is possible to make the sum of elements of $$$a$$$ equal to $$$0$$$, and \"NO\" otherwise.

You can output the answer in any case (upper or lower). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be recognized as positive responses.

Example
Input
5
1
-1
2
1 -1
2
1 1
5
1 -1 1 -1 1
6
-1 1 -1 -1 -1 -1
Output
NO
YES
NO
NO
YES
Note

In the first example, we cannot perform any operations, and the sum of elements of $$$a$$$ equals $$$-1 \\neq 0$$$, so the answer is \"NO\".

In the second example, the sum of elements of $$$a$$$ is already $$$0$$$, so the answer is \"YES\".

In the fifth example, we can perform an operation with $$$i = 3$$$, transforming $$$a$$$ as follows: $$$[-1, 1, \\color{red}{-1, -1}, -1, -1] \\rightarrow [-1, 1, \\color{red}{1, 1}, -1, -1]$$$. The sum of the resulting array is $$$0$$$, so the answer is \"YES\".

\n\n\n\nSource lookup status:\n\n- Editorial (en): https://codeforces.com/blog/entry/155337 loaded (200 OK; title: \"Codeforces Round 1111 (Div. 2) Editorial - Codeforces\"; mentions 2247A)\n\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A deep editorial explaining like literally everything.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be JUST the hint text. No \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nint main() {\n setIO();\n}\n```" } ] }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:06:26.992Z" }, { "type": "assistant", "uuid": "ca6bf846-cee4-4d79-bcf2-55602e29e61a", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "model": "claude-opus-5", "id": "msg_011CdV4Jb9wEg8fGhy1qhJJn", "type": "message", "role": "assistant", "content": [ { "type": "thinking", "thinking": "", "signature": 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If so, what is the solution/condition given?" }, "caller": { "type": "direct" } } ], "stop_reason": "tool_use", "stop_sequence": null, "stop_details": null, "usage": { "input_tokens": 2, "cache_creation_input_tokens": 4874, "cache_read_input_tokens": 0, "output_tokens": 2581, "server_tool_use": { "web_search_requests": 0, "web_fetch_requests": 0 }, "service_tier": "standard", "cache_creation": { "ephemeral_1h_input_tokens": 4874, "ephemeral_5m_input_tokens": 0 }, "inference_geo": "not_available", "iterations": [ { "input_tokens": 2, "output_tokens": 2581, "cache_read_input_tokens": 0, "cache_creation_input_tokens": 4874, "cache_creation": { "ephemeral_5m_input_tokens": 0, "ephemeral_1h_input_tokens": 4874 }, "type": "message" } ], "speed": "standard" }, "diagnostics": null }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:06:54.632Z" }, { "type": "user", "uuid": "636f1089-e9f5-409b-a544-f5b27b66cd27", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "role": "user", "content": [ { "tool_use_id": "toolu_01EJvoL4Se9cQ4qVVaPL2xQY", "type": "tool_result", "content": "The server returned HTTP 403 Forbidden.\n\nThe response body was not retrieved. If this URL requires authentication, use an authenticated tool (e.g. `gh` for GitHub, or an MCP-provided fetch tool) instead of WebFetch." } ] }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:06:54.776Z" }, { "type": "assistant", "uuid": "d86e92c8-d16a-495c-8ab8-8e4b41a68eeb", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "model": "claude-opus-5", "id": "msg_011CdV4Jb9wEg8fGhy1qhJJn", "type": "message", "role": "assistant", "content": [ { "type": "tool_use", "id": "toolu_019r8ujb9PKi8MxuPDejp9Vu", "name": "WebSearch", "input": { "query": "Codeforces 2247A \"Zero Sum\" editorial sum divisible by 4" }, "caller": { "type": "direct" } } ], "stop_reason": "tool_use", "stop_sequence": null, "stop_details": null, "usage": { "input_tokens": 2, "cache_creation_input_tokens": 4874, "cache_read_input_tokens": 0, "output_tokens": 2581, "server_tool_use": { "web_search_requests": 0, "web_fetch_requests": 0 }, "service_tier": "standard", "cache_creation": { "ephemeral_1h_input_tokens": 4874, "ephemeral_5m_input_tokens": 0 }, "inference_geo": "not_available", "iterations": [ { "input_tokens": 2, "output_tokens": 2581, "cache_read_input_tokens": 0, "cache_creation_input_tokens": 4874, "cache_creation": { "ephemeral_5m_input_tokens": 0, "ephemeral_1h_input_tokens": 4874 }, "type": "message" } ], "speed": "standard" }, "diagnostics": null }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:06:55.030Z" }, { "type": "user", "uuid": "609f0216-fee4-4ffa-ad62-0f2b0f78b9bd", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "role": "user", "content": [ { "tool_use_id": "toolu_019r8ujb9PKi8MxuPDejp9Vu", "type": "tool_result", "content": "Web search results for query: \"Codeforces 2247A \"Zero Sum\" editorial sum divisible by 4\"\n\nLinks: [{\"title\":\"Codeforces Round 1111 (Div. 2) Editorial - Codeforces\",\"url\":\"https://codeforces.com/blog/entry/155337\"},{\"title\":\"Problem - 2247A - Codeforces\",\"url\":\"https://codeforces.com/problemset/problem/2247/A\"},{\"title\":\"https://codeforces.com/problemset/status/2247/prob...\",\"url\":\"https://codeforces.com/problemset/status/2247/problem/A\"},{\"title\":\"Codeforces Global Round 23 Editorial - Codeforces\",\"url\":\"https://codeforces.com/blog/entry/108011\"},{\"title\":\"Editorial of Educational Codeforces Round 7 - Codeforces\",\"url\":\"https://codeforces.com/blog/entry/23442\"},{\"title\":\"Problem - A - Codeforces\",\"url\":\"https://codeforces.com/contest/2247/problem/A\"},{\"title\":\"Educational Codeforces Round 77 Editorial - Codeforces\",\"url\":\"https://codeforces.com/blog/entry/71805\"}]\n\nI found the editorial for Codeforces 2247A \"Zero Sum\". Here's the key information regarding the sum divisible by 4:\n\nThe key insight is that the sum of all elements modulo 4 does not change after a single operation. \n\nTherefore, if the sum of all elements modulo 4 is not equal to 0, then the answer is \"NO\".\n\nThe solution approach is:\n1. First check if the initial sum is divisible by 4 (i.e., sum mod 4 ≠ 0 means the answer is \"NO\")\n2. If the sum is divisible by 4, the solution constructs the array alternately with values 1, -1, 1, etc., by performing operations on indices 1 through n-1 in order\n3. After this construction, the absolute value of the sum becomes at most 2, so it must equal zero since zero is the only number divisible by 4 in that range\n\nThis elegant solution relies on the invariant that operations preserve the sum modulo 4.\n\n\nREMINDER: You MUST include the sources above in your response to the user using markdown hyperlinks." } ] }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:07:00.881Z" }, { "type": "assistant", "uuid": "829ffff4-552a-40a8-b8f0-9ce4d9257374", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "model": "claude-opus-5", "id": "msg_011CdV4M5ckhZkGt3egeLrwU", "type": "message", "role": "assistant", "content": [ { "type": "thinking", "thinking": "", "signature": 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"msg_011CdV4M5ckhZkGt3egeLrwU", "type": "message", "role": "assistant", "content": [ { "type": "tool_use", "id": "toolu_0155kiHW3e7AZVP97vbjnMsh", "name": "StructuredOutput", "input": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "Don't try to search over reachable arrays — that space is huge. Instead, ask the classic question for \"can I reach state X with these moves\": **what quantity does one operation fail to change?** Start by literally enumerating the four cases of the pair $(a_i, a_{i+1})$ you might hit: $(1,1)$, $(1,-1)$, $(-1,1)$, $(-1,-1)$, and compute how the total sum changes in each." }, { "order": 2, "content": "Do the case work:\n\n- $(1,1) \\to (-1,-1)$: sum drops by $4$.\n- $(-1,-1) \\to (1,1)$: sum rises by $4$.\n- $(1,-1)$ or $(-1,1)$: they just swap places sign-wise, sum unchanged.\n\nSo every operation changes the sum by $0$, $+4$, or $-4$. Nothing else. What does that tell you about $S \\bmod 4$ where $S = \\sum a_i$?" }, { "order": 3, "content": "$S \\bmod 4$ is an **invariant** — no sequence of operations can ever touch it. Since the target sum is $0$ and $0 \\bmod 4 = 0$, a necessary condition is $S \\equiv 0 \\pmod 4$.\n\nAlso note this quietly handles parity for you: $S \\equiv n \\pmod 2$ always, so $4 \\mid S$ already forces $n$ to be even. You don't need a separate \"n must be even\" check. The real question left: is $4 \\mid S$ also *sufficient*?" }, { "order": 4, "content": "Yes, and here's a clean way to prove it: **normalize the array**. Sweep $i = 1, 2, \\ldots, n-1$; at step $i$, if $a_i$ isn't the value you want (say you want the pattern $1, -1, 1, -1, \\ldots$), apply the operation at index $i$ — that fixes $a_i$ forever and only disturbs $a_{i+1}$, which you haven't locked yet.\n\nAfter the sweep, $a_1..a_{n-1}$ perfectly alternate and only $a_n$ is free, so the final sum is in $\\{-1, 0, 1, 2\\}$-ish territory — in any case $|S| \\le 2$. Combine that with the invariant: the sum is still $\\equiv 0 \\pmod 4$ and has absolute value $\\le 2$, so it must be exactly $0$." }, { "order": 5, "content": "The whole answer is: print **YES** iff $\\left(\\sum_i a_i\\right) \\bmod 4 = 0$.\n\nOne C++ gotcha: `%` on negatives gives negative results (`-2 % 4 == -2`, but `-1 % 4 == -1`, and `-4 % 4 == 0`), so `sum % 4 == 0` actually still works fine — but `abs(sum) % 4 == 0` is the bulletproof version. $O(n)$ per test, no DP, no number theory beyond \"mod 4\"." } ], "editorial": "## TL;DR\n\nAnswer is `YES` if and only if $4 \\mid \\sum_{i=1}^{n} a_i$. That's it. One loop, one modulo, done. The `dp` and `number theory` tags on this problem are a bit of a troll — the only \"number theory\" here is the number $4$.\n\n## Setting up\n\nWe have $a$ of length $n$ with $a_i \\in \\{-1, 1\\}$. The move: pick $i \\le n-1$ and flip **both** $a_i$ and $a_{i+1}$. Question: can we reach total sum $0$?\n\nThe state space is $2^n$ arrays, and $n \\le 100$, so brute force is dead on arrival. Whenever a problem asks \"is state $X$ reachable under these moves\", the move that basically always works is: **find an invariant** — some function of the array that no operation can change. If the invariant of the start differs from the invariant of the target, the answer is instantly NO. Then you separately argue that when the invariants match, you can actually get there.\n\n## Step 1: what does one operation do to the sum?\n\nLet $S = \\sum_i a_i$. The operation only touches $a_i$ and $a_{i+1}$, so only their contribution to $S$ can change. Four cases:\n\n| $(a_i, a_{i+1})$ | after flip | contribution before | after | $\\Delta S$ |\n|---|---|---|---|---|\n| $(1, 1)$ | $(-1,-1)$ | $+2$ | $-2$ | $-4$ |\n| $(-1,-1)$ | $(1,1)$ | $-2$ | $+2$ | $+4$ |\n| $(1,-1)$ | $(-1,1)$ | $0$ | $0$ | $0$ |\n| $(-1,1)$ | $(1,-1)$ | $0$ | $0$ | $0$ |\n\nSo **every operation changes $S$ by exactly $0$, $+4$, or $-4$.** Therefore\n\n$$S \\bmod 4 \\text{ is invariant.}$$\n\nEquivalent framing if you like counting instead: let $k$ = number of $-1$s. Then $S = n - 2k$. An operation changes $k$ by $+2$, $-2$, or $0$, so **the parity of $k$ is invariant**. Reaching $S = 0$ means reaching $k = n/2$, which needs $n$ even and $k \\equiv n/2 \\pmod 2$. Grind through the algebra and you get the exact same condition, $4 \\mid S$. Pick whichever version your brain likes.\n\n## Step 2: necessity\n\nTarget sum is $0$, and $0 \\equiv 0 \\pmod 4$. Since $S \\bmod 4$ never changes, if the initial $S$ is not divisible by $4$, the answer is **NO**. Done.\n\nA nice free bonus: $S \\equiv n \\pmod 2$ always (each of the $n$ terms is odd). So $4 \\mid S \\Rightarrow S$ even $\\Rightarrow n$ even. You do **not** need a separate \"is $n$ even?\" check — it's baked in. (Odd $n$ always gives odd $S$, which is never divisible by $4$.)\n\n## Step 3: sufficiency — the construction\n\nNecessary conditions are cheap; the interesting half is showing $4 \\mid S$ is *enough*. Two proofs, both short.\n\n### Proof A: the left-to-right normalization sweep\n\nHere's the key structural fact about this operation: **applying it at index $i$ changes $a_i$ and $a_{i+1}$, but nothing to the left of $i$.** So we can process positions left to right and \"lock in\" whatever value we want, one at a time.\n\nTarget the alternating pattern $1, -1, 1, -1, \\ldots$. For $i = 1, 2, \\ldots, n-1$: if $a_i$ isn't the value the pattern demands, apply the operation at $i$. This fixes $a_i$ (and never touches it again, since all future operations are at indices $\\ge i+1$), at the cost of flipping $a_{i+1}$ — which we haven't locked yet, so who cares.\n\nAfter the sweep, $a_1, \\ldots, a_{n-1}$ are exactly $1, -1, 1, -1, \\ldots$, and $a_n$ is whatever it ended up as. Now compute:\n\n- If $n$ is **even**: $a_1..a_{n-1}$ has an odd count $n-1$ of alternating terms starting at $1$, summing to $1$. Total $S \\in \\{1 + 1, 1 - 1\\} = \\{2, 0\\}$.\n- If $n$ is **odd**: $a_1..a_{n-1}$ has an even count, summing to $0$. Total $S \\in \\{1, -1\\}$.\n\nIn every case $|S| \\le 2$ after normalization. But the invariant says $S \\equiv 0 \\pmod 4$ still holds, and the only multiple of $4$ with absolute value $\\le 2$ is $0$. So $S = 0$. **The condition is sufficient, and this sweep is an explicit constructive recipe** if the problem ever asked you to print the moves.\n\n### Proof B: pigeonhole descent\n\nSuppose $S > 0$ (so more $1$s than $-1$s) and $4 \\mid S$, hence $S \\ge 4$. Claim: there must exist two **adjacent** $1$s. Why? If no two $1$s were adjacent, every $1$ would need a $-1$ separating it from the next $1$, giving $\\#(-1) \\ge \\#(1) - 1$, i.e. $S = \\#(1) - \\#(-1) \\le 1$. Contradiction with $S \\ge 4$.\n\nSo find an adjacent pair of $1$s, flip it, and $S$ drops by exactly $4$. Repeat. Symmetrically, if $S < 0$ you always find two adjacent $-1$s and raise $S$ by $4$. Each step moves $S$ by $4$ toward zero while keeping $S \\equiv 0 \\pmod 4$, so you land exactly on $0$. Same conclusion, different flavor.\n\n## Step 4: the algorithm\n\n```\nread n, read array\nS = sum of array\nprint \"YES\" if S % 4 == 0 else \"NO\"\n```\n\nComplexity: $O(n)$ time, $O(1)$ memory per test case. With $t \\le 200$ and $n \\le 100$ this is roughly 20k operations total — the I/O is more expensive than the \"algorithm\".\n\n## Checking against the samples\n\n| Test | array | $S$ | $S \\bmod 4$ | answer |\n|---|---|---|---|---|\n| 1 | $[-1]$ | $-1$ | $\\ne 0$ | NO ✓ |\n| 2 | $[1,-1]$ | $0$ | $0$ | YES ✓ |\n| 3 | $[1,1]$ | $2$ | $\\ne 0$ | NO ✓ |\n| 4 | $[1,-1,1,-1,1]$ | $1$ | $\\ne 0$ | NO ✓ |\n| 5 | $[-1,1,-1,-1,-1,-1]$ | $-4$ | $0$ | YES ✓ |\n\nAll five match.\n\n## Traps people actually fall into\n\n- **\"$n$ even $\\Rightarrow$ YES.\"** Nope. Sample 3 is $n = 2$, $[1, 1]$, sum $2$: you can only turn it into $[-1,-1]$, sum $-2$. Forever stuck. Even length is necessary but nowhere near sufficient.\n- **Modulo of negative numbers.** In C++, `-1 % 4` is `-1`, not `3`. Luckily the test `sum % 4 == 0` is unaffected (a negative multiple of $4$ still gives $0$), but if you ever write `sum % 4 == 0 || sum % 4 == ...` style logic, or compare against a positive residue, you'll get burned. Using `abs(sum) % 4 == 0` removes all doubt.\n- **Reaching for DP.** With $n \\le 100$ someone will absolutely try a reachability DP over \"sum so far after committing flips\". It even works, since the flip pattern reduces to choosing a subset of indices to toggle — but it's like bringing a chainsaw to cut a birthday cake. The invariant argument is three lines and actually proves the answer.\n- **Thinking order of operations matters.** It doesn't, in the sense that applying the operation at the same index twice cancels out, so the reachable set is determined purely by which indices you use an odd number of times. That's another route to the same result: if $x_i \\in \\{0,1\\}$ says whether index $i$ is used oddly, then $a_j$ ends up flipped iff $x_{j-1} + x_j$ is odd — and the number of flipped positions is always even (it's the number of \"boundaries\" in the $x$ sequence, counting $x_0 = x_n = 0$). Flipping an even number of $\\pm 1$ entries changes the sum by a multiple of... $4$? Not quite — flipping $2m$ entries changes $S$ by $-2\\sum_{\\text{flipped}} a_j$, and that sum has the same parity as $2m$, i.e. it's even, so $\\Delta S$ is a multiple of $4$. Consistent, as it must be.\n\n## The one-line takeaway\n\nAdjacent-pair sign flips preserve $\\sum a_i \\bmod 4$, and that's the *only* obstruction — so `4 | sum` is exactly the answer.", "solution": "#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\n/*\n * Flipping a_i and a_{i+1} changes the total sum by:\n * (1, 1) -> (-1,-1) : -4\n * (-1,-1) -> (1, 1) : +4\n * (1,-1) / (-1,1) : 0\n * So sum mod 4 is invariant => 4 | sum is necessary.\n *\n * It is also sufficient: sweep i = 1..n-1 and use the operation at i to force\n * a_i into the alternating pattern 1,-1,1,-1,... (this never touches indices\n * < i). Afterwards only a_n is free, so |sum| <= 2; being still divisible by 4\n * forces sum == 0.\n */\nint main() {\n setIO();\n\n int t;\n cin >> t;\n while (t--) {\n int n;\n cin >> n;\n int sum = 0;\n for (int i = 0; i < n; i++) {\n int x;\n cin >> x;\n sum += x;\n }\n cout << (abs(sum) % 4 == 0 ? \"YES\" : \"NO\") << '\\n';\n }\n return 0;\n}" }, "caller": { "type": "direct" } } ], "stop_reason": "tool_use", "stop_sequence": null, "stop_details": null, "usage": { "input_tokens": 2, "cache_creation_input_tokens": 3452, "cache_read_input_tokens": 4874, "output_tokens": 5957, "server_tool_use": { "web_search_requests": 0, "web_fetch_requests": 0 }, "service_tier": "standard", "cache_creation": { "ephemeral_1h_input_tokens": 3452, "ephemeral_5m_input_tokens": 0 }, "inference_geo": "not_available", "iterations": [ { "input_tokens": 2, "output_tokens": 5957, "cache_read_input_tokens": 4874, "cache_creation_input_tokens": 3452, "cache_creation": { "ephemeral_5m_input_tokens": 0, "ephemeral_1h_input_tokens": 3452 }, "type": "message" } ], "speed": "standard" }, "diagnostics": null }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:08:10.853Z" }, { "type": "user", "uuid": "b94bbddc-0a54-4faa-8637-72d896aa72ab", "session_id": "da60faa1-91a5-46fa-beb5-d42e3fd86ec4", "message": { "role": "user", "content": [ { "tool_use_id": "toolu_0155kiHW3e7AZVP97vbjnMsh", "type": "tool_result", "content": "Structured output provided successfully" } ] }, "parent_tool_use_id": null, "parent_agent_id": null, "timestamp": "2026-07-28T20:08:10.861Z" } ]