**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
// Problem: H. 378QAQ and Core
// Contest: Codeforces - Codeforces Round 947 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1975/problem/H
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#include <chrono>
std::mt19937 eng(std::chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return std::uniform_int_distribution<int>(l, r)(eng); }

namespace FastIO {
	template <typename T> inline T read() { T x = 0, w = 0; char ch = getchar(); while (ch < '0' || ch > '9') w |= (ch == '-'), ch = getchar(); while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar(); return w ? -x : x; }
	template <typename T> inline void write(T x) { if (!x) return; write<T>(x / 10), putchar((x % 10) ^ '0'); }
	template <typename T> inline void print(T x) { if (x > 0) write<T>(x); else if (x < 0) putchar('-'), write<T>(-x); else putchar('0'); }
	template <typename T> inline void print(T x, char en) { print<T>(x), putchar(en); }
	inline char rChar() { char ch = getchar(); while (!isalpha(ch)) ch = getchar(); return ch; }
}; using namespace FastIO;

#define MAXN 1000001
char s[MAXN], t[MAXN], r[MAXN];

const int Z = 26;
int cnt[Z], a[MAXN];
void solve() {
	int N = read<int>(), peg = 0; scanf("%s", s + 1);
	for (int i = 0; i < Z; ++i) cnt[i] = 0;
	for (int i = 1; i <= N; ++i) ++cnt[s[i] - 'a'];
	for (int i = 0, len = 0, cur; i < Z; ++i) if (cnt[i] > 0) {
		// cnt[i] elements, len + 1 spaces.
		int d = cnt[i] % (len + 1), w = cnt[i] / (len + 1);
		for (int i = 0; i <= len; ++i) a[i] = w;
		if (d > 0) {
			int L = 1, R = d;
			while (L < R) {
				int mid = L + R >> 1;
				if ((len + 1) / (mid + 1) * mid + (len + 1) % (mid + 1) >= d) R = mid; else L = mid + 1;
			}
			for (int i = len, t = d; ~i && t; --i)
				if ((len - i + 1) % (L + 1) != 0) --t, ++a[i];
		}
		peg = a[len], cur = 0;
		while (a[0]--) r[++cur] = 'a' + i;
		for (int j = 1; j <= len; ++j) {
			r[++cur] = t[j]; while (a[j]--) r[++cur] = 'a' + i;
		}
		len += cnt[i];
		for (int j = 1; j <= len; ++j) t[j] = r[j];
	}
	char ch = 'z'; while (cnt[ch - 'a'] == 0) --ch;
	for (int i = 1, j = 1; i <= N; i = ++j) {
		while (j <= N && t[j] == ch) ++j;
		if (j - i >= peg) {
			for (int k = i; k <= N; ++k) putchar(t[k]); puts("");
			return;
		}
	}
}

int main() { int T = read<int>(); while (T--) solve(); return 0; }

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases where the code might fail. The main issue in the code seems to be the logic for distributing characters to form the lexicographically minimum core. Let's break down the potential edge cases:

1. **Single Character Strings**: These are straightforward and should not expose bugs.
2. **Strings with All Identical Characters**: These should also be straightforward.
3. **Strings with a Single Unique Character**: This is a bit more complex, as the distribution logic might fail if not handled correctly.
4. **Strings with a Mix of Characters**: This is where the distribution logic can go wrong.

Given the complexity, let's focus on a mix of characters that can challenge the distribution logic. Specifically, a string with a few unique characters and a large number of a single character can be tricky.

### Test Input
We will create a string where the majority of the characters are the same, but a few unique characters are interspersed. This can expose issues in the distribution logic.

```plaintext
1
10
aaabccz
```

### Explanation
- **Test Case**: 1
- **String Length**: 10
- **String**: "aaabccz"

### Expected Output
The lexicographically minimum possible core for this string should be "a" because the string can be rearranged to "aabczzz" and the core of "aabczzz" is "a".

### Why This Test Input?
- **Majority of One Character**: The string has 6 'a's, which are the majority.
- **Few Unique Characters**: The string also has 'b', 'c', and 'z', which are fewer in number.
- **Distribution Logic**: The distribution logic in the code might not handle the distribution of 'a's correctly when there are a few unique characters.

This test input should expose any issues in the code's logic for distributing characters to form the lexicographically minimum core.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include <bits/stdc++.h>
#include <immintrin.h>

using namespace std;

using ll = long long;
using ld = long double;
#define x first
#define y second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

mt19937 mt(123);

void solve();
void init();

int32_t main() {
#ifndef LOCAL
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    cout << fixed << setprecision(30);
    init();
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

void init() {}

bool cmp(const vector<int>& a, const vector<int>& b) {
    for (int i = 0; i < min(a.size(), b.size()); i++) {
        if (a[i] < b[i]) {
            return true;
        }
        if (b[i] > a[i]) {
            return false;
        }
    }
    if (a.size() > b.size()) {
        return true;
    }
    return false;
}

vector<int> slv(vector<int> cnt) {
    // for (int i = 0; i < cnt.size(); i++) {
    //     cout << cnt[i] << ' ';
    // }
    // cout << '\n';
    vector<int> ans;
    if (cnt.back() == 1) {
        // for (int i = 0; i < cnt.size(); i++) {
        //     for (int j = 0; j < cnt[i]; j++) {
        //         ans.push_back(i);
        //     }
        // }
        ans.push_back(cnt.size() - 1);
        return ans;
    }
    if (cnt.size() == 1) {
        return vector<int>(cnt[0], 0);
    }
    int n = 0;
    for (int i = 0; i < cnt.size(); i++) {
        n += cnt[i];
    }
    int mx = cnt.back();
    if (n - mx < mx - 1) {
        vector<int> nwcnt = cnt;
        nwcnt.pop_back();
        nwcnt.push_back(mx - (n - mx) - 1);
        vector<int> res = slv(nwcnt);
        for (int i = 0; i < res.size(); i++) {
            ans.push_back(cnt.size() - 1);
            if (res[i] != cnt.size() - 1) {
                ans.push_back(res[i]);
            }
        }
        ans.push_back(cnt.size() - 1);
        return ans;
    }
    mx--;
    vector<vector<int>> nw(mx);
    cnt.pop_back();
    int nwpos = 0, nwsz = mx, nwnwsz = -1;
    while (cnt.back() != 0) {
        reverse(all(nw));
        nwnwsz = nwsz;
        for (int j = 0; j < nwsz && cnt.back() != 0; j++) {
            if (cnt[nwpos] == 0) {
                nwpos++;
                nwnwsz = min(j + 1, nwnwsz);
            }
            cnt[nwpos]--;
            nw[j].push_back(nwpos);
        }
        nwsz = nwnwsz;
        sort(all(nw), cmp);
    }
    vector<int> nw_cnt(1, 1);
    for (int i = 1; i < nw.size(); i++) {
        if (nw[i] == nw[i - 1]) {
            nw_cnt.back()++;
        } else {
            nw_cnt.push_back(1);
        }
    }
    vector<int> res = slv(nw_cnt);
    // cout << "! ";
    // for (int i : res) {
    //     cout << i << ' ';
    // }
    // cout << '\n';
    // cout << nw[0].size() << '\n';
    for (int i : res) {
        ans.push_back(cnt.size());
        for (int j : nw[i]) {
            ans.push_back(j);
        }
    }
    ans.push_back(cnt.size());
    return ans;
}

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<int> cnt(26, 0), norm_cnt, num;
    for (char c : s) {
        cnt[c - 'a']++;
    }
    // while (cnt.back() == 0) {
    //     cnt.pop_back();
    // }
    for (int i = 0; i < 26; i++) {
        if (cnt[i]) {
            num.push_back(i);
            norm_cnt.push_back(cnt[i]);
        }
    }
    vector<int> ans = slv(norm_cnt);
    for (int i = 0; i < ans.size(); i++) {
        ans[i] = num[ans[i]];
    }
    for (int i : ans) {
        cout << char('a' + i);
    }
    cout << '\n';
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
```

6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include <list>
#include <vector>
#include <assert.h>
#include <iostream>
#include <algorithm>
using namespace std;
struct str
{
    list<char> d;
    void pushfront(str s) { reverse(begin(s.d), end(s.d)); for (char i : s.d) d.push_front(i); }
    void pushback (str s) { for (char i : s.d) d.push_back(i); }
    void show() const { for (char i : d) cout << i; }
    bool operator != (str s) { return d != s.d; }
};

void solve(vector<pair<str, int>> &v, const str &s, const int k)
{
    if (empty(v))
    {
        for (int i(1); i <= k; ++i) s.show();
        return;
    }
    if (k == 1) return s.show();
    int a(0);
    for (auto &[i, j] : v) a += j;
    // cout << v.size() << ' '; s.show(); cout << ' ' << k << endl;
    if (a < k)
    {
        for (auto &[i, j] : v) i.pushfront(s);
        if (k - 1 > a) solve(v, s, k - 1 - a);
        else
        {
            const str t(v.back().first); a = v.back().second;
            v.pop_back(); solve(v, t, a);
        }
        return s.show();
    }
    vector<str> c(k - 1, s);
    int l(0), p(0);
    for (auto [i, j] : v)
    {
        l = p;
        while (j--)
        {
            c[p].pushback(i);
            if (++p == k - 1) p = l;
        }
    }
    v.clear();
    for (str i : c)
        if (empty(v) || i != v.back().first) v.push_back({i, 1});
        else ++v.back().second;
    sort(begin(v), end(v), [](const pair<str, int> x, const pair<str, int> y) -> bool
                           {
                            //    for (int i(0); i < (int)size(x.first.d) && i < (int)size(y.first.d); ++i)
                            //        if      (x.first.d[i] < y.first.d[i]) return true;
                            //        else if (x.first.d[i] > y.first.d[i]) return false;
                               auto i(begin(x.first.d)), j(begin(y.first.d));
                               while (i != end(x.first.d) && j != end(y.first.d))
                               {
                                if (*i < *j) return true;
                                if (*i > *j) return false;
                               }
                                return size(x.first.d) > size(y.first.d);
                            //    return size(x.first.d) > size(y.first.d);
                           });
    const str t(v.back().first); a = v.back().second;
    v.pop_back(); solve(v, t, a); s.show();
}

int main()
{
    // freopen("test.in", "r", stdin);
    // freopen("test.out", "w", stdout);
    int t; cin >> t;
    while (t--)
    {
        int n, a[26] = {}; vector<pair<str, int>> v; string s; cin >> n >> s;
        for (char i : s) ++a[i - 'a'];
        for (int i(0); i < 26; ++i) if (a[i]) { str t; t.d.push_back(i + 'a'); v.push_back({t, a[i]}); }
        const str c(v.back().first); const int k(v.back().second); v.pop_back();
        solve(v, c, k); cout << endl;
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
abc
4
abcd
6
abcabc
6
zzzzzz
7
ababccc
7
ccbabcc
```

6
3
abc
4
abcd
6
abcabc
6
zzzzzz
7
ababccc
7
ccbabcc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
from itertools import permutations

def find_min_core(s):
    n = len(s)
    min_core = None

    # Generate all permutations of the string s
    for perm in permutations(s):
        perm_str = ''.join(perm)

        # Find the maximum lexicographic substring (core)
        max_suffix = ''
        for i in range(n):
            suffix = perm_str[i:]
            if suffix > max_suffix:
                max_suffix = suffix

        # Update the minimum core found so far
        if min_core is None or max_suffix < min_core:
            min_core = max_suffix

    return min_core

# Read input
import sys
input = sys.stdin.read
data = input().split()

# Parse the input
t = int(data[0])
index = 1
results = []

for _ in range(t):
    n = int(data[index])
    s = data[index + 1]
    index += 2

    # Find and store the result for this test case
    result = find_min_core(s)
    results.append(result)

# Output all results
for result in results:
    print(result)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
1
a
2
ab
3
abc
4
abac
5
ababc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
from itertools import permutations

def find_smallest_largest_substring(s):
    smallest_largest_substring = None

    # Generate each permutation on-the-fly and find the largest substring for each
    for p in permutations(s):
        permuted_str = ''.join(p)
        largest_substring = max(
            permuted_str[x:y] for x in range(len(permuted_str)) for y in range(x + 1, len(permuted_str) + 1))

        if smallest_largest_substring is None or largest_substring < smallest_largest_substring:
            smallest_largest_substring = largest_substring

    return smallest_largest_substring

n = int(input())
for i in range(n):
    l = int(input())
    s = input()
    result = find_smallest_largest_substring(s)
    print(result)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
abcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
from itertools import permutations
from itertools import combinations
n = int(input())
for i in range(n):
    l = int(input())
    s = input()
    perms = [''.join(p) for p in permutations(s)]
    permutation = set(perms)
    l1 = []
    for j in permutation:
        res = [j[x:y] for x, y in combinations(
            range(len(j) + 1), r=2)]
        res.sort()
        l1 += [res[-1]]
    l1.sort()
    print(l1[0])

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
aaa
4
abcd
6
zzzzzz
6
ababab
7
abcabcz
7
ccccccc
```

### Explanation:
1. **Case 1: "aaa"**
   - This is a simple case with all characters the same. The core should be "aaa" as it is the only possible substring.

2. **Case 2: "abcd"**
   - This case tests a string with all distinct characters. The core should be "abcd" as it is the only possible maximum lexicographic substring.

3. **Case 3: "zzzzzz"**
   - Similar to the first case, all characters are the same. The core should be "zzzzzz".

4. **Case 4: "ababab"**
   - This case tests a string with repeated characters. The core should be "ababab" as it is the only possible maximum lexicographic substring.

5. **Case 5: "abcabcz"**
   - This case tests a string with a mix of characters and a single unique character at the end. The core should be "z" as it is the lexicographically maximum substring.

6. **Case 6: "ccccccc"**
   - This case tests a string with all characters the same. The core should be "ccccccc".

These test cases are designed to cover various edge cases and ensure that the buggy code is tested thoroughly.

6
3
aaa
4
abcd
6
zzzzzz
6
ababab
7
abcabcz
7
ccccccc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder out = new StringBuilder();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            if (t == 100000 && ig >= 2500) continue;
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.append(vals[n]).append("\n");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (j > n) {
                    start = nextStart;
                    j = start;
                } else if (i >= start) {
                    start = j;
                    nextStart = start;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != start && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            int v = n;
            while (v > 0) {
                out.append(vals[v]);
                v = nexts[v];
            }
            out.append("\n");
        }

        System.out.println(out);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code might fail. The code appears to handle the lexicographically minimum core calculation by sorting the characters and then trying to rearrange them to find the minimum core. However, there are a few potential issues:

1. **Handling of repeated characters**: The code might not correctly handle cases where there are many repeated characters.
2. **Edge cases with single or few characters**: The code might not handle strings with very few unique characters correctly.
3. **Complex rearrangements**: The code might fail to correctly identify the minimum core when the string has a complex arrangement of characters.

Let's consider a small-scale test input that can expose these issues:

1. **String with all identical characters**: This should be straightforward, but it's a good edge case to check.
2. **String with two different characters**: This can test the code's ability to handle simple rearrangements.
3. **String with a mix of characters**: This can test the code's ability to handle more complex rearrangements.

Here is a test input that covers these scenarios:

```plaintext
3
5
aaaaa
4
abba
6
abacda
```

Explanation:
1. **"aaaaa"**: This string has all identical characters. The core should be "a".
2. **"abba"**: This string has two different characters. The core should be "a".
3. **"abacda"**: This string has a mix of characters. The core should be "a".

This test input is small-scale and complete, and it should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder out = new StringBuilder();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            if (t == 100000) continue;
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.append(vals[n]).append("\n");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (j > n) {
                    start = nextStart;
                    j = start;
                } else if (i >= start) {
                    start = j;
                    nextStart = start;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != start && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            int v = n;
            while (v > 0) {
                out.append(vals[v]);
                v = nexts[v];
            }
            out.append("\n");
        }

        System.out.println(out);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that might cause the code to fail. The buggy code has a conditional check `if (t == 100000) continue;` which skips processing if the number of test cases is 100,000. This suggests that the code might have issues with large input sizes or specific patterns in the input.

Given the constraints:
- \(1 \leq t \leq 10^5\)
- \(1 \leq n \leq 10^6\)
- The sum of \(n\) over all test cases does not exceed \(10^6\)

We should create a test case that:
1. Has a small number of test cases to ensure it is not skipped.
2. Includes a string that tests the core logic of the algorithm, especially the part where the code might fail due to incorrect handling of the core substring.

Let's create a test case with a single test case and a string that has a repeating character pattern, which can help expose issues in the logic for finding the lexicographically minimum core.

```plaintext
1
5
abcba
```

This test case has:
- \(t = 1\) (one test case)
- \(n = 5\) (the length of the string)
- The string "abcba"

The expected output for this test case should be "abcba" because "abcba" is the only possible rearrangement of the string "abcba" that maintains the same core.

This test case is small and complete, and it should help expose any issues in the core logic of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
vector<char>s[1000005],Now[1000005],res,res1,t[1000005];
int n,ct[26];
char str[1000005];
int Ct[1000005],tt;
bool cmp(int a,int b){
	if(s[a].size()!=s[b].size())return 0;int len=s[a].size();
	for(int i=0;i<len;++i)if(s[a][i]!=s[b][i])return 0;
	return 1;
}
bool operator<(vector<char>&a,vector<char>&b){
	int l=min(a.size(),b.size());
	for(int i=0;i<l;++i){
		if(a[i]<b[i])return 1;
		if(a[i]>b[i])return 0;
	}return (a.size()<b.size());
}
void ins(vector<char>&a,vector<char>&b){for(auto v:b)a.pb(v);}
void calc(){
	res.clear();
	res1.clear();
	while(1){
		int ct=0;
		for(int i=n;i>=1;--i)if(cmp(i,n))++ct;
		if(ct==1){
			res=s[n];
         	for(int i=1;i<n;++i)if(s[i]<res1)ins(res,s[i]);else break;
         	break;
		}
		if(ct>=n-ct+1){
			if(ct==n){
				for(int i=1;i<=n;++i)ins(res,s[i]);
				break;
			}
			int ct1=n-ct;++ct1;
			for(int i=1;i<ct1;++i)Now[i]=s[i];
			reverse(Now+1,Now+ct1);
			vector<char>lst1=s[n];
			n=ct1;
			int a=(ct+ct1-1)/ct1,b=ct%ct1;
			if(!b){
				reverse(Now+1,Now+ct1);
				for(int i=1;i<ct1;++i){
					s[i].clear();
				    for(int j=1;j<=a;++j)ins(s[i],lst1);
				    ins(s[i],Now[i]);
			    }--n;
			    reverse(lst1.begin(),lst1.end());
			    for(int j=1;j<=a;++j)ins(res1,lst1);
			    continue;
			}
			for(int i=1;i<=ct1-b;++i){
				s[i].clear();
				for(int j=1;j<a;++j)ins(s[i],lst1);
				ins(s[i],Now[i]);
			}
			reverse(Now+ct1-b+1,Now+ct1);
			for(int i=1;i<b;++i){
				s[ct1-i].clear();
				for(int j=1;j<=a;++j)ins(s[ct1-i],lst1);
				ins(s[ct1-i],Now[i]);
			}
			if(b==1&&!(s[1]<res)){
				for(int i=1;i<=a;++i)ins(res,lst1);
				break;
			}
			--n,reverse(lst1.begin(),lst1.end());
			for(int j=1;j<=a;++j)ins(res1,lst1);
			continue;
		}
		int len=s[n].size();
		for(int i=len-1;i>=0;--i)res1.pb(s[n][i]);
		--n,--ct;
		tt=0;int nn=n-ct;
		for(int i=1;i<=nn;){
			int r=i;while(r<nn&&cmp(r,r+1))++r;
			++tt;t[tt]=s[i];Ct[tt]=r-i+1,i=r+1;
		}
		vector<char>lst=s[n];n=ct;
		for(int i=1;i<=n;++i)s[i]=lst;
		int now=1,ct1=n;
		while(1){
			while(now<=tt&&!Ct[now])++now;
			if(now>tt)break;
			if(ct1<=Ct[now]){
				Ct[now]-=ct1;
				for(int i=n-ct1+1;i<=n;++i)ins(s[i],t[now]);
				continue;
			}
			for(int i=n-ct1+1;i<=n-ct1+Ct[now];++i)ins(s[i],t[now]);
			ct1-=Ct[now],Ct[now]=0;
		}

	}
	reverse(res1.begin(),res1.end());
	ins(res,res1);
}
void solve(){
	scanf("%d",&n);
	memset(ct,0,sizeof(ct));scanf("%s",str+1);
	for(int i=1;i<=n;++i)++ct[(int)(str[i]-'a')];n=0;
	for(int i=0;i<26;++i){while(ct[i]--)++n,s[n].clear(),s[n].pb((char)(i+'a'));}
	calc();
	for(auto v:res)putchar(v);
	putchar('\n');
}
int main(){
	int t;cin>>t;
	while(t--)solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
abc
4
abca
6
abcabc
6
zzzzzz
7
ababbbb
7
ccbabcc
```

6
3
abc
4
abca
6
abcabc
6
zzzzzz
7
ababbbb
7
ccbabcc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
vector<char>s[1000005],Now[1000005],res,res1,t[1000005];
int n,ct[26];
char str[1000005];
int Ct[1000005],tt;
bool cmp(int a,int b){
	if(s[a].size()!=s[b].size())return 0;int len=s[a].size();
	for(int i=0;i<len;++i)if(s[a][i]!=s[b][i])return 0;
	return 1;
}
bool operator<(vector<char>&a,vector<char>&b){
	int l=min(a.size(),b.size());
	for(int i=0;i<l;++i){
		if(a[i]<b[i])return 1;
		if(a[i]>b[i])return 0;
	}return (a.size()<b.size());
}
void ins(vector<char>&a,vector<char>&b){for(auto v:b)a.pb(v);}
void calc(){
	res.clear();
	res1.clear();
	while(1){
		int ct=0;
		for(int i=n;i>=1;--i)if(cmp(i,n))++ct;
		if(ct==1){
			res=s[n];reverse(res1.begin(),res1.end());
         	for(int i=1;i<n;++i)if(s[i]<res1)ins(res,s[i]);else break;
         	reverse(res1.begin(),res1.end());
         	break;
		}
		if(ct>=n-ct+1){
			if(ct==n){
				for(int i=1;i<=n;++i)ins(res,s[i]);
				break;
			}
			int ct1=n-ct;++ct1;
			for(int i=1;i<ct1;++i)Now[i]=s[i];
			reverse(Now+1,Now+ct1);
			vector<char>lst1=s[n];
			n=ct1;
			int a=(ct+ct1-1)/ct1,b=ct%ct1;
			if(!b){
				reverse(Now+1,Now+ct1);
				for(int i=1;i<ct1;++i){
					s[i].clear();
				    for(int j=1;j<=a;++j)ins(s[i],lst1);
				    ins(s[i],Now[i]);
			    }--n;
			    reverse(lst1.begin(),lst1.end());
			    for(int j=1;j<=a;++j)ins(res1,lst1);
			    continue;
			}
			for(int i=1;i<=ct1-b;++i){
				s[i].clear();
				for(int j=1;j<a;++j)ins(s[i],lst1);
				ins(s[i],Now[i]);
			}
			reverse(Now+ct1-b+1,Now+ct1);
			for(int i=1;i<b;++i){
				s[ct1-i].clear();
				for(int j=1;j<=a;++j)ins(s[ct1-i],lst1);
				ins(s[ct1-i],Now[ct1-i]);
			}
			reverse(res1.begin(),res1.end());
			if(b==1&&!(s[1]<res1)){
				for(int i=1;i<=a;++i)ins(res,lst1);
				reverse(res1.begin(),res1.end());
				break;
			}
			reverse(res1.begin(),res1.end());
			--n,reverse(lst1.begin(),lst1.end());
			for(int j=1;j<=a;++j)ins(res1,lst1);
			continue;
		}
		int len=s[n].size();
		for(int i=len-1;i>=0;--i)res1.pb(s[n][i]);
		--n,--ct;
		tt=0;int nn=n-ct;
		for(int i=1;i<=nn;){
			int r=i;while(r<nn&&cmp(r,r+1))++r;
			++tt;t[tt]=s[i];Ct[tt]=r-i+1,i=r+1;
		}
		vector<char>lst=s[n];n=ct;
		for(int i=1;i<=n;++i)s[i]=lst;
		int now=1,ct1=n;
		while(1){
			while(now<=tt&&!Ct[now])++now;
			if(now>tt)break;
			if(ct1<=Ct[now]){
				Ct[now]-=ct1;
				for(int i=n-ct1+1;i<=n;++i)ins(s[i],t[now]);
				continue;
			}
			for(int i=n-ct1+1;i<=n-ct1+Ct[now];++i)ins(s[i],t[now]);
			ct1-=Ct[now],Ct[now]=0;
		}
	}
	reverse(res1.begin(),res1.end());
	ins(res,res1);
}
void solve(){
	scanf("%d",&n);
	memset(ct,0,sizeof(ct));scanf("%s",str+1);
	for(int i=1;i<=n;++i)++ct[(int)(str[i]-'a')];n=0;
	for(int i=0;i<26;++i){while(ct[i]--)++n,s[n].clear(),s[n].pb((char)(i+'a'));}
	calc();
	for(auto v:res)putchar(v);
	putchar('\n');
}
int main(){
	int t;cin>>t;
	while(t--)solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
abc
4
abcd
6
abcabc
6
zzzzzz
7
ababccc
7
cbabccc
```

6
3
abc
4
abcd
6
abcabc
6
zzzzzz
7
ababccc
7
cbabccc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder out = new StringBuilder();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            if (t == 100000 && ig >= 100) continue;
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.append(vals[n]).append("\n");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (j > n) {
                    start = nextStart;
                    j = start;
                } else if (i >= start) {
                    start = j;
                    nextStart = start;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != start && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            int v = n;
            while (v > 0) {
                out.append(vals[v]);
                v = nexts[v];
            }
            out.append("\n");
        }

        System.out.println(out);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
abc
4
abcc
6
abcabc
6
zzzzzz
7
ababbbb
7
ccbabcc
```

6
3
abc
4
abcc
6
abcabc
6
zzzzzz
7
ababbbb
7
ccbabcc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder out = new StringBuilder();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            if (t == 100000 && ig >= 50000) continue;
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.append(vals[n]).append("\n");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (j > n) {
                    start = nextStart;
                    j = start;
                } else if (i >= start) {
                    start = j;
                    nextStart = start;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != start && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            int v = n;
            while (v > 0) {
                out.append(vals[v]);
                v = nexts[v];
            }
            out.append("\n");
        }

        System.out.println(out);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
a
2
ab
3
aba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
//    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        StringBuilder out = new StringBuilder();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.append(vals[n]).append("\n");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (j > n) {
                    start = nextStart;
                    j = start;
                } else if (i >= start) {
                    start = j;
                    nextStart = start;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != start && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            int v = n;
            while (v > 0) {
                out.append(vals[v]);
                v = nexts[v];
            }
            out.append("\n");
        }

        System.out.println(out);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with handling certain rearrangements and edge cases, particularly when the string has multiple characters with the same frequency or when the characters are already in a specific order.

Let's consider the following steps:

1. **Single Character Strings**: Test cases with a single character to ensure the code handles them correctly.
2. **All Characters the Same**: Test cases where all characters are the same to check if the code can handle such uniform strings.
3. **Mixed Characters with High Frequency**: Test cases with mixed characters where some characters have high frequencies to see how the code rearranges them.
4. **Strings with Consecutive Duplicates**: Test cases with consecutive duplicates to check if the code can handle such patterns.
5. **Strings with Lexicographically Challenging Patterns**: Test cases with strings that have a mix of characters to challenge the lexicographic ordering.

Here is a small-scale test input that covers these scenarios:

```plaintext
5
1
a
4
cccc
6
aabbcc
6
ababab
6
abcabc
```

### Explanation:
1. **Single Character String**:
   - `1 a` - The only possible core is "a".

2. **All Characters the Same**:
   - `4 cccc` - The only possible core is "cccc".

3. **Mixed Characters with High Frequency**:
   - `6 aabbcc` - The lexicographically minimum core should be "ababcc" or any permutation that results in the same core.

4. **Strings with Consecutive Duplicates**:
   - `6 ababab` - The lexicographically minimum core should be "ababab".

5. **Strings with Lexicographically Challenging Patterns**:
   - `6 abcabc` - The lexicographically minimum core should be "abcabc".

This test input should help expose any issues in the buggy code, particularly with handling different character frequencies and lexicographic ordering.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
import java.util.*;
import java.io.*;

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args)  throws IOException {
        int t = nextInt();
        for (int ig = 0; ig < t; ++ig) {
            int n = nextInt();
            String s = next();
            int[] cnts = new int[26];
            for (int i = 0; i < n; ++i) cnts[s.charAt(i) - 'a']++;
            char[] vals = new char[n + 1];
            vals[0] = (char)('a' - 1);
            int ind = 1;
            for (int i = 0; i < 26; ++i) {
                while (cnts[i]-- > 0) vals[ind++] = (char)('a' + i);
            }
            if (vals[n] != vals[n - 1]) {
                out.println(vals[n] + "");
                continue;
            }
            int[] nexts = new int[n + 1];
            Arrays.fill(nexts, -1);
            int start = 0;
            int[] lasts = new int[n + 1];
            for (int i = 0; i <= n; ++i) lasts[i] = i;
            boolean[] changes = new boolean[n + 1];
            for (int i = 1; i <= n; ++i) {
                changes[i] = vals[i] != vals[i - 1];
                if (changes[i]) start = i;
            }
            int nextStart = start;
            for (int i = 0, j = start ; i < n;) {
                if (i >= start) {
                    if (j > n) {
                        start = nextStart;
                        j = start;
                    } else {
                        start = j;
                    }
                } else if (j > n) {
                    j = nextStart;
                } else {
                    nexts[lasts[j]] = i;
                    lasts[j] = lasts[i];
                    changes[j] = j != nextStart && changes[i];
                    if (changes[j]) nextStart = j;
                    ++i;
                    ++j;
                }
            }

            StringBuilder sb = new StringBuilder();
            int v = n;
            while (v > 0) {
                sb.append(vals[v]);
                v = nexts[v];
            }
            out.println(sb);
        }

        out.flush();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
3
aaa
4
abcd
5
abaca
6
abcabc
7
ababccc
```

5
3
aaa
4
abcd
5
abaca
6
abcabc
7
ababccc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(3)
#define DB double
#define LL long long
#define ULL unsigned long long
#define in128 __int128
#define cint const int
#define cLL const LL
#define For(z,e1,e2) for(int z=(e1);z<=(e2);z++)
#define Rof(z,e1,e2) for(int z=(e2);z>=(e1);z--)
#define For_(z,e1,e2) for(int z=(e1);z<(e2);z++)
#define Rof_(z,e1,e2) for(int z=(e2);z>(e1);z--)
#define inint(e) scanf("%d",&e)
#define inll(e) scanf("%lld",&e)
#define inpr(e1,e2) scanf("%d%d",&e1,&e2)
#define in3(e1,e2,e3) scanf("%d%d%d",&e1,&e2,&e3)
#define outint(e) printf("%d\n",e)
#define outint_(e) printf("%d%c",e," \n"[i==n])
#define outint2_(e,e1,e2) printf("%d%c",e," \n"[(e1)==(e2)])
#define outll(e) printf("%lld\n",e)
#define outll_(e) printf("%lld%c",e," \n"[i==n])
#define outll2_(e,e1,e2) printf("%lld%c",e," \n"[(e1)==(e2)])
#define exc(e) if(e) continue
#define stop(e) if(e) break
#define ret(e) if(e) return
#define ll(e) (1ll*(e))
#define pb push_back
#define ft first
#define sc second
#define pii pair<int,int>
#define pli pair<LL,int>
#define vct vector
#define clean(e) while(!e.empty()) e.pop()
#define all(ev) ev.begin(),ev.end()
#define sz(ev) ((int)ev.size())
#define debug(x) printf("%s=%d\n",#x,x)
#define x0 __xx00__
#define y1 __yy11__
#define ffo fflush(stdout)
// cLL mod=998244353,G=404;
cLL mod=1e9+7;
// cLL mod[2]={1686688681ll,1888686661ll},base[2]={166686661ll,188868881ll};
template <typename Type> void get_min(Type &w1,const Type w2) { if(w2<w1) w1=w2; } template <typename Type> void get_max(Type &w1,const Type w2) { if(w2>w1) w1=w2; }
template <typename Type> Type up_div(Type w1,Type w2) { return (w1/w2+(w1%w2?1:0)); }
template <typename Type> Type gcd(Type X_,Type Y_) { Type R_=X_%Y_; while(R_) { X_=Y_; Y_=R_; R_=X_%Y_; } return Y_; } template <typename Type> Type lcm(Type X_,Type Y_) { return (X_/gcd(X_,Y_)*Y_); }
template <typename Type> Type md(Type w1,const Type w2=mod) { w1%=w2; if(w1<0) w1+=w2; return w1; } template <typename Type> Type md_(Type w1,const Type w2=mod) { w1%=w2; if(w1<=0) w1+=w2; return w1; }
void ex_gcd(LL &X_,LL &Y_,LL A_,LL B_) { if(!B_) { X_=1ll; Y_=0ll; return ; } ex_gcd(Y_,X_,B_,A_%B_); X_=md(X_,B_); Y_=(1ll-X_*A_)/B_; } LL inv(LL A_,LL B_=mod) { LL X_=0ll,Y_=0ll; ex_gcd(X_,Y_,A_,B_); return X_; }
template <typename Type> void add(Type &w1,const Type w2,const Type M_=mod) { w1=md(w1+w2,M_); } void mul(LL &w1,cLL w2,cLL M_=mod) { w1=md(w1*md(w2,M_),M_); } template <typename Type> Type pw(Type X_,Type Y_,Type M_=mod) { Type S_=1; while(Y_) { if(Y_&1) mul(S_,X_,M_); Y_>>=1; mul(X_,X_,M_); } return S_; }
template <typename Type> Type bk(vector <Type> &V_) { auto T_=V_.back(); V_.pop_back(); return T_; } template <typename Type> Type tp(stack <Type> &V_) { auto T_=V_.top(); V_.pop(); return T_; } template <typename Type> Type frt(queue <Type> &V_) { auto T_=V_.front(); V_.pop(); return T_; }
template <typename Type> Type bg(set <Type> &V_) { auto T_=*V_.begin(); V_.erase(V_.begin()); return T_; } template <typename Type> Type bk(set <Type> &V_) { auto T_=*prev(V_.end()); V_.erase(*prev(V_.end())); return T_; }
mt19937 gen(time(NULL)); int rd() { return abs((int)gen()); }
void main_init()
{

}
cint N=1.02e6;
char s[N];
void solve(vct <int> &v)
{
    // for(auto i:v) printf("%d ",i);
    // printf("\n");
    // ffo;
    int n=sz(v),m=v.back();
    if(n==1||v[n-2]!=m)
    {
        v={v.back()};
        return ;
    }
    int cnt=n-1;
    while(cnt>0&&v[cnt-1]==m) cnt--;
    v.resize(cnt);
    cnt=n-cnt-1;
    n-=cnt+1;
    vct <vct<int>> g(1,(vct<int>){});
    vct <int> v2;
    int h=cnt,tot=0,lst=0,w=0;
    for(int l=0,r=0;l<n;l=(++r))
    {
        while(r+1<n&&v[r+1]==v[r]) r++;
        int len=r-l+1,c=v[l];
        Rof(k,1,len/h) g[w].pb(c);
        len%=h;
        exc(!len);
        tot++;
        lst+=len;
        g.pb(g.back());
        g[w++].pb(c);
        h-=len;
        while(len)
        {
            len--;
            v2.pb(tot);
        }
    }
    tot++;
    Rof(i,1,cnt-lst) v2.pb(tot);
    // reverse(all(g));
    v.clear(); vct <int> emp; swap(v,emp);
    solve(v2);
    v.pb(m);
    for(auto i:v2)
    {
        i--;
        for(auto c:g[i])
            v.pb(c);
        v.pb(m);
    }
}
void main_solve()
{
    int n;
    scanf("%d%s",&n,s);
    vct <int> v(n,0);
    For_(i,0,n) v[i]=s[i]-'a'+1;
    sort(all(v));
    solve(v);
    for(auto i:v) printf("%c",'a'+i-1);
    printf("\n");
}
int main()
{
    // ios::sync_with_stdio(0); cin.tie(0);
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    // srand(time(NULL));
    main_init();
    int _; inint(_); For(__,1,_) // T>1 ?
        // printf("\n------------\n\n"),
        main_solve();
    return 0;
}
/*

*/

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 378QAQ and Core
time_limit_ms: 2000
memory_limit_mb: 256
Description: 378QAQ has a string $s$ of length $n$. Define the core of a string as the substring$^\dagger$ with maximum lexicographic$^\ddagger$ order.

For example, the core of "$\mathtt{bazoka}$" is "$\mathtt{zoka}$", and the core of "$\mathtt{aaa}$" is "$\mathtt{aaa}$".

378QAQ wants to rearrange the string $s$ so that the core is lexicographically minimum. Find the lexicographically minimum possible core over all rearrangements of $s$.

$^\dagger$ A substring of string $s$ is a continuous segment of letters from $s$. For example, "$\mathtt{defor}$", "$\mathtt{code}$" and "$\mathtt{o}$" are all substrings of "$\mathtt{codeforces}$" while "$\mathtt{codes}$" and "$\mathtt{aaa}$" are not.

$^\ddagger$ A string $p$ is lexicographically smaller than a string $q$ if and only if one of the following holds:

  * $p$ is a prefix of $q$, but $p \ne q$; or    * in the first position where $p$ and $q$ differ, the string $p$ has a smaller element than the corresponding element in $q$ (when compared by their ASCII code).

For example, "$\mathtt{code}$" and "$\mathtt{coda}$" are both lexicographically smaller than "$\mathtt{codeforces}$" while "$\mathtt{codeforceston}$" and "$\mathtt{z}$" are not.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\leq t\leq 10^5$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1\leq n\leq 10^6$) — the length of string $s$.

The next line of each test case contains the string $s$ of length $n$. The string $s$ consists of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, output the lexicographically minimum possible core over all rearrangements of $s$.

Examples:
input:
6
3
qaq
4
cccc
6
bazoka
6
zazzzz
7
ababbbb
7
ccbabcc
output:
qaq
cccc
z
zzz
bbababb
cbcacbc

Note: In the first test case, all possible rearrangements and their corresponding cores are as follows:

  * "$\mathtt{qaq}$", its core is "$\mathtt{qaq}$".    * "$\mathtt{aqq}$", its core is "$\mathtt{qq}$".    * "$\mathtt{qqa}$", its core is "$\mathtt{qqa}$".

So the core with the minimum lexicographic order in all rearrangement plans is "$\mathtt{qaq}$".

Buggy Code:
//RCh - H
#include<bits/stdc++.h>
#define INF LONG_LONG_MAX
#define inf INT_MAX
#define bctz __builtin_ctz
#define bclz __builtin_clz
#define bpt __builtin_popcount
#define pb push_back
#define x first
#define y second

using namespace std;

typedef long long INT;
typedef pair<int, int> pii;
typedef pair<int, pii> pi3;

template <class T> inline void smax(T &a, T b){if(a < b)a = b;}
template <class T> inline void smin(T &a, T b){if(a > b)a = b;}

const int NN = 2e5 + 10;
const int mod = 998244353;

int A[NN], B[NN];
char s[NN];
string ans;

void solve(vector <string> &vec){
	reverse(vec.begin(), vec.end());
	string mx = vec[0];
	vector <string> buf;
	while(vec.back() != mx){
		buf.pb(vec.back());
		vec.pop_back();
	}
	vec.pop_back();
	int n = vec.size(), m = buf.size();
	if(n == 0){
		string tmp;
		for(string u : buf) tmp += u;
		tmp += mx;
		vec.pb(tmp);
		ans = mx;
		return;
	}
	int pos = 0, now = 0;
	for(int i = 0; i < m; i++){
		vec[now++] += buf[i];
		if(now != n && i < m - 1 && buf[i] != buf[i + 1]) pos = now;
		if(now == n) now = pos;
	}
	solve(vec);
	vec[0] += mx;
	if(!ans.empty()) ans += mx;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	int t;
	scanf("%d", &t);
	while(t--){
		int n;
		scanf("%d", &n);
		scanf("%s", s + 1);
		sort(s + 1, s + n + 1);
		vector <string> vec;
		string tmp;
		for(int i = 1; i <= n; i++){
			tmp.clear();
			tmp += s[i];
			vec.pb(tmp);
		}
		solve(vec);
		cout << ans << endl;
		ans.clear();
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6
3
abc
4
abcc
6
abccba
6
zzzzzz
7
ababccc
7
ccbabcc
```

6
3
abc
4
abcc
6
abccba
6
zzzzzz
7
ababccc
7
ccbabcc