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          "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 2250B: \"String Construction\".\n\nKnown problem metadata:\n- Contest ID: 2250\n- Index: B\n- Name: String Construction\n- Rating: unknown / unrated\n- Tags: constructive algorithms\n- Problem URL: https://codeforces.com/contest/2250/problem/B\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">B. String Construction</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>1 second</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p> </p><p>You are given two integers $$$n$$$ and $$$k$$$.</p><p>Construct a binary string$$$^{\\text{∗}}$$$ $$$s$$$ of length $$$n$$$, such that both of the following conditions hold:</p><ul> <li> The absolute difference between the number of characters $$$\\mathtt{0}$$$ and the number of characters $$$\\mathtt{1}$$$ in $$$s$$$ is at most $$$1$$$. </li><li> There are exactly $$$k$$$ pairs of adjacent equal characters in $$$s$$$. Formally, there are exactly $$$k$$$ indices $$$i$$$ ($$$1 \\le i \\le n-1$$$) satisfying $$$s_i = s_{i + 1}$$$. </li></ul><p>Or determine that no such string exists.</p><div class=\"statement-footnote\"><p>$$$^{\\text{∗}}$$$A binary string is a string where each character is either $$$\\mathtt{0}$$$ or $$$\\mathtt{1}$$$.</p></div></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 1000$$$). The description of the test cases follows.</p><p>The only line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$, $$$0 \\le k \\le n-1$$$).</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output a binary string $$$s$$$ of length $$$n$$$ — the string you constructed. Print $$$-1$$$ if such a string does not exist.</p><p>If there are multiple answers, you may output any of them.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">8</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">5 2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">4 3</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">6 1</div><div class=\"test-example-line test-example-line-even test-example-line-4\">5 0</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">7 3</div><div class=\"test-example-line test-example-line-even test-example-line-6\">4 2</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">3 2</div><div class=\"test-example-line test-example-line-even test-example-line-8\">7 4</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">01110</div><div class=\"test-example-line test-example-line-even test-example-line-2\">-1</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">101001</div><div class=\"test-example-line test-example-line-even test-example-line-4\">01010</div><div class=\"test-example-line test-example-line-odd test-example-line-5\">0100011</div><div class=\"test-example-line test-example-line-even test-example-line-6\">0011</div><div class=\"test-example-line test-example-line-odd test-example-line-7\">-1</div><div class=\"test-example-line test-example-line-even test-example-line-8\">0111000</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first test case, one possible answer is $$$s=\\mathtt{01110}$$$. It contains three characters $$$\\mathtt{1}$$$ and two characters $$$\\mathtt{0}$$$, and there are exactly $$$2$$$ adjacent equal pairs in $$$s$$$: $$$(s_2, s_3)$$$ and $$$(s_3, s_4)$$$.</p><p>In the second test case, $$$k=n-1$$$. All characters in $$$s$$$ should be equal, so the numbers of characters $$$\\mathtt{0}$$$ and $$$\\mathtt{1}$$$ could not differ by at most $$$1$$$. Thus, the answer is $$$-1$$$.</p><p>In the third test case, note that $$$\\mathtt{010110}$$$ is also a possible answer.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/155516 loaded (200 OK; title: \"Codeforces Round 1112 (Div. 1, Div. 2) Editorial - Codeforces\"; mentions 2250B)\n</source-lookup-status>\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 <bits/stdc++.h>\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```"
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      "content": [
        {
          "type": "text",
          "text": "Solved it straight from the statement — this one's a classic \"stop thinking about characters, start thinking about blocks\" problem."
        }
      ],
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            "status": "success",
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            "hints": [
              {
                "order": 1,
                "content": "Stop staring at individual characters and chop the string into **maximal runs** (blocks) of equal characters. For example $\\mathtt{0001101}$ splits into $\\mathtt{000}\\mid\\mathtt{11}\\mid\\mathtt{0}\\mid\\mathtt{1}$. Now count adjacent-equal pairs in terms of those blocks instead of positions."
              },
              {
                "order": 2,
                "content": "A block of length $L$ contributes exactly $L-1$ adjacent equal pairs, and pairs at block boundaries contribute nothing. So if the string has $b$ blocks with lengths $L_1+\\dots+L_b=n$, the total is $\\sum (L_i-1) = n-b$. Meaning: $k$ doesn't leave you any freedom — it *pins down* the number of blocks to $b = n-k$."
              },
              {
                "order": 3,
                "content": "Blocks strictly alternate $\\mathtt{0},\\mathtt{1},\\mathtt{0},\\mathtt{1},\\dots$ by definition (otherwise two neighboring blocks would merge). So out of $b$ blocks, one character owns $\\lceil b/2\\rceil$ blocks and the other owns $\\lfloor b/2\\rfloor$. Every block needs at least $1$ character. The whole question becomes: can you hand out $n$ characters so that the zero-count and one-count differ by at most $1$ while each side covers its block minimums?"
              },
              {
                "order": 4,
                "content": "Take zeros $=\\lceil n/2\\rceil$ and ones $=\\lfloor n/2\\rfloor$ (perfectly balanced by construction). Since $b=n-k\\le n$, we automatically get $\\lceil b/2\\rceil\\le\\lceil n/2\\rceil$ and $\\lfloor b/2\\rfloor\\le\\lfloor n/2\\rfloor$, so the minimums are always satisfiable... except for one degenerate case: if $b=1$ the second character gets **zero** blocks, forcing its count to be exactly $0$, and $|n-0|\\le 1$ is false for $n\\ge2$. So the answer is $-1$ if and only if $k=n-1$. Parity of $k$ or $n$ is irrelevant — don't invent extra impossible cases."
              },
              {
                "order": 5,
                "content": "Construction: give every block length $1$ (that's the alternating string with $b$ blocks), then dump the leftover zeros into the *first* $\\mathtt{0}$-block and the leftover ones into the *first* $\\mathtt{1}$-block. Concretely with $b=n-k$, $b_0=\\lceil b/2\\rceil$, $b_1=\\lfloor b/2\\rfloor$, $n_0=\\lceil n/2\\rceil$, $n_1=\\lfloor n/2\\rfloor$, print $\\underbrace{0\\cdots0}_{n_0-b_0+1}\\underbrace{1\\cdots1}_{n_1-b_1+1}0\\,1\\,0\\,1\\cdots$ until $b$ blocks total. Careful: dumping *all* leftovers on a single character breaks the balance condition — split them between the two."
              }
            ],
            "editorial": "## The reframe that kills the problem\n\nCounting \"indices $i$ with $s_i = s_{i+1}$\" position by position is a trap. Split $s$ into **maximal runs** (blocks) of equal characters:\n\n$$\\mathtt{0001101} \\;\\to\\; \\mathtt{000}\\mid\\mathtt{11}\\mid\\mathtt{0}\\mid\\mathtt{1}$$\n\nA block of length $L$ contributes exactly $L-1$ equal-adjacent pairs (inside it), and every boundary between two consecutive blocks contributes $0$ (characters differ there — that's what \"maximal\" means). So if there are $b$ blocks with lengths $L_1,\\dots,L_b$ summing to $n$:\n\n$$k \\;=\\; \\sum_{i=1}^{b}(L_i - 1) \\;=\\; n - b.$$\n\nEquivalently: there are $n-1$ adjacent positions total, exactly $b-1$ of them are \"switch points\", so equal pairs $=(n-1)-(b-1)=n-b$. Either way, the same punchline:\n\n> **$k$ doesn't give you a choice, it fixes the block count: $b = n - k$.**\n\nThat's the whole problem. Everything else is bookkeeping.\n\n## When is it impossible?\n\nBlocks alternate characters — forced, by maximality. So with $b$ blocks, one character owns $b_0=\\lceil b/2\\rceil$ blocks and the other owns $b_1=\\lfloor b/2\\rfloor$. Let's say the string starts with $\\mathtt{0}$, so $\\mathtt{0}$ owns $b_0$ blocks and $\\mathtt{1}$ owns $b_1$.\n\nEach block needs at least one character, so we need counts $n_0 \\ge b_0$ and $n_1 \\ge b_1$ with $n_0+n_1=n$ and $|n_0-n_1|\\le 1$. The tightest balanced split is $n_0=\\lceil n/2\\rceil$, $n_1=\\lfloor n/2\\rfloor$. Since $b=n-k \\le n$ (because $k\\ge0$), monotonicity of ceil/floor gives\n\n$$\\left\\lceil \\tfrac b2\\right\\rceil \\le \\left\\lceil \\tfrac n2\\right\\rceil, \\qquad \\left\\lfloor \\tfrac b2\\right\\rfloor \\le \\left\\lfloor \\tfrac n2\\right\\rfloor,$$\n\ni.e. $n_0 \\ge b_0$ and $n_1 \\ge b_1$ **always hold**. So a valid string exists for basically every input...\n\n...with exactly one exception. If $b=1$, then $b_1=0$: the second character owns *no* blocks, which forces $n_1 = 0$ exactly (not \"$\\ge 0$\" — a character that never appears appears zero times). Then $|n_0-n_1| = n \\ge 2 > 1$. Dead.\n\n$b=1 \\iff k=n-1$, which matches the samples $(4,3)$ and $(3,2)$ giving $-1$.\n\n**Answer: print $-1$ if and only if $k = n-1$. Otherwise a string always exists.**\n\nLet me pre-emptively murder some false assumptions people bring to this problem:\n\n- *\"Parity of $k$ must match something.\"* Nope. $n=4,k=2 \\to \\mathtt{0011}$. $n=4,k=1 \\to \\mathtt{0110}$... wait, that's $b=3$: $\\mathtt{0}\\mid\\mathtt{11}\\mid\\mathtt{0}$, one pair. Both parities are fine.\n- *\"Large $k$ is hard.\"* The only hard $k$ is the maximum one, $n-1$. Even $k=n-2$ is fine ($b=2$: half zeros, half ones, e.g. $\\mathtt{000111}$).\n- *\"$n$ odd is a problem.\"* It isn't; $|n_0-n_1|\\le1$ explicitly allows the off-by-one.\n\n## The construction\n\nWe need $b=n-k$ blocks, alternating, with $\\lceil n/2\\rceil$ zeros and $\\lfloor n/2\\rfloor$ ones. Start from the skeleton where **every block has length 1** — that's just the alternating string $\\mathtt{0101\\ldots}$ truncated to $b$ characters, which uses $b_0$ zeros and $b_1$ ones. Now we have leftovers:\n\n$$e_0 = \\left\\lceil \\tfrac n2\\right\\rceil - b_0 \\ge 0, \\qquad e_1 = \\left\\lfloor \\tfrac n2\\right\\rfloor - b_1 \\ge 0,$$\n\nand $e_0+e_1 = n-b = k$. Stuff all the leftover zeros into the first $\\mathtt{0}$-block and all the leftover ones into the first $\\mathtt{1}$-block. The result:\n\n$$s \\;=\\; \\underbrace{\\mathtt{0\\cdots0}}_{e_0+1}\\;\\underbrace{\\mathtt{1\\cdots1}}_{e_1+1}\\;\\mathtt{0}\\,\\mathtt{1}\\,\\mathtt{0}\\,\\mathtt{1}\\cdots \\quad(\\text{$b$ blocks total}).$$\n\nWhy this is correct, checked against both conditions:\n\n1. **Block count is still $b$** — we only grew existing blocks, never created or merged any (growing block 1 keeps it adjacent to a different character, since block 2 is the opposite character). Hence pairs $=n-b=k$. ✅\n2. **Counts** are $\\lceil n/2\\rceil$ zeros and $\\lfloor n/2\\rfloor$ ones by construction, difference $\\le 1$. ✅\n3. **Well-defined**: since $b\\ge2$ there really is a first $\\mathtt{1}$-block to dump $e_1$ into, and $b_1 \\ge 1$, $n_1=\\lfloor n/2\\rfloor\\ge1$ for $n\\ge2$.\n\nThe one trap here: don't be lazy and dump *all* $k$ extras into a single block. For $n=6,k=3$ that gives $\\mathtt{0000}\\ldots$ and the balance condition eats you alive. Split the surplus between the two characters — the \"make each side exactly balanced first, then compute per-character surplus\" framing does that automatically.\n\n### Worked examples\n\n- $n=5,k=2$: $b=3$, $b_0=2,b_1=1$, $n_0=3,n_1=2$, so $e_0=1,e_1=1$ → $\\mathtt{00}\\,\\mathtt{11}\\,\\mathtt{0} = \\mathtt{00110}$. Zeros 3, ones 2, pairs at $(1,2)$ and $(3,4)$: exactly 2. ✅ (Sample prints $\\mathtt{01110}$; any valid answer is accepted.)\n- $n=7,k=4$: $b=3$, $b_0=2,b_1=1$, $n_0=4,n_1=3$, $e_0=2,e_1=2$ → $\\mathtt{000\\,111\\,0}$. Pairs $=7-3=4$. ✅\n- $n=6,k=1$: $b=5$, $b_0=3,b_1=2$, $n_0=n_1=3$, $e_0=0,e_1=1$ → $\\mathtt{0\\,11\\,0\\,1\\,0} = \\mathtt{011010}$. One pair. ✅\n- $n=5,k=0$: $b=5=n$, $e_0=e_1=0$ → $\\mathtt{01010}$. ✅\n\n## Complexity\n\n$O(n)$ time and $O(n)$ memory per test case, $O(\\sum n)$ overall — trivially inside limits for $\\sum n \\le 2\\cdot10^5$. Only real performance note: with $t$ up to $1000$ and total output up to $2\\cdot10^5$ characters, build the answer into one buffer (or at least use fast I/O) instead of flushing per test. `endl` is not your friend.\n\n## Reference implementation sketch\n\n```\nb  = n - k\nif b == 1: print -1\nb0 = (b+1)/2, b1 = b/2\nn0 = (n+1)/2, n1 = n/2\ne0 = n0 - b0, e1 = n1 - b1\nemit block i (0-indexed) of char ('0' if i even else '1')\n  with length 1 + (e0 if i==0 else e1 if i==1 else 0)\n```\n\nShort, no case analysis beyond the single $-1$ check, and provably correct for every input in range.",
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