**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cards Partition
time_limit_ms: 2000
memory_limit_mb: 256
Description: [DJ Genki vs Gram - Einherjar Joker](https://soundcloud.com/leon- hwang-368077289/einherjar-joker-dj-genki-vs-gram)

You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy at most $k$ new cards in total, and the cards you buy can contain any integer between $\mathbf{1}$ and $\mathbf{n}$, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

  * all the decks must have the same size;    * there are no pairs of cards with the same value in the same deck.

Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^{16}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{10}$, $\sum a_i \geq 1$) — the number of cards of type $i$ you have at the beginning, for each $1 \leq i \leq n$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Examples:
input:
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
output:
2
3
1
7
2
2
1
1
2

Note: In the first test case, you can buy one card with the number $1$, and your cards become $[1, 1, 1, 1, 2, 2, 3, 3]$. You can partition them into the decks $[1, 2], [1, 2], [1, 3], [1, 3]$: they all have size $2$, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than $2$, so the answer is $2$.

In the second test case, you can buy two cards with the number $1$ and one card with the number $3$, and your cards become $[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5]$, which can be partitioned into $[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 5], [2, 3, 5], [2, 4, 5]$. You can show that you cannot get a partition with decks of size greater than $3$, so the answer is $3$.

Buggy Code:
import time
import sys
from math import inf, log2
input = sys.stdin.readline
from bisect import bisect_left, bisect_right, insort_left, insort_right
mod = 1000000007
from collections import deque, Counter, defaultdict as dd
from heapq import heappop, heappush, heapify
from itertools import accumulate
from functools import lru_cache
inty=lambda: int(input())
stringy=lambda: input().strip()
normal=lambda: input().split()
mappy=lambda: map(int,input().strip().split())
fmappy=lambda: map(float,input().strip().split())
listy=lambda: list(map(int,input().strip().split()))
#sys.setrecursionlimit(10**5 + 10000)
tick=lambda: time.perf_counter()
logg=lambda a, b: log2(a) / log2(b)
acc=lambda x: list(accumulate(x))
sign=lambda x, y: x >= 0 and y >= 0 or x < 0 and y < 0

class Solution:
    def solution(self, n, k, a):
        maxi, m = max(a), sum(a)
        if (maxi * n) - m <= k: return n
        for s in range(n, 0, -1):
            if k >= (maxi * s) - m >= 0 or (m > (maxi * s) and (m % s == 0 or (m + k) % s == 0)):
                return s

sol = Solution()
T = inty()
for _ in range(T):
    n, k = mappy()
    a = listy()
    print(sol.solution(n, k, a))

**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
3 2
1 1 1
4 3
1 2 2 3
5 5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cards Partition
time_limit_ms: 2000
memory_limit_mb: 256
Description: [DJ Genki vs Gram - Einherjar Joker](https://soundcloud.com/leon- hwang-368077289/einherjar-joker-dj-genki-vs-gram)

You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy at most $k$ new cards in total, and the cards you buy can contain any integer between $\mathbf{1}$ and $\mathbf{n}$, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

  * all the decks must have the same size;    * there are no pairs of cards with the same value in the same deck.

Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^{16}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{10}$, $\sum a_i \geq 1$) — the number of cards of type $i$ you have at the beginning, for each $1 \leq i \leq n$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Examples:
input:
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
output:
2
3
1
7
2
2
1
1
2

Note: In the first test case, you can buy one card with the number $1$, and your cards become $[1, 1, 1, 1, 2, 2, 3, 3]$. You can partition them into the decks $[1, 2], [1, 2], [1, 3], [1, 3]$: they all have size $2$, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than $2$, so the answer is $2$.

In the second test case, you can buy two cards with the number $1$ and one card with the number $3$, and your cards become $[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5]$, which can be partitioned into $[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 5], [2, 3, 5], [2, 4, 5]$. You can show that you cannot get a partition with decks of size greater than $3$, so the answer is $3$.

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

using namespace std;
#define ll                long long
#define yes               cout<<"YES\n"
#define no                cout<<"NO\n"
#define FAST              (ios_base::sync_with_stdio(false),cin.tie(NULL))
#define sz(x)             (int)(x).size()

int main()
{
   FAST;
    int t;
    cin>>t;

    while(t--){
        int n,k;
        cin>>n>>k;
        ll maxi = INT_MIN;
        ll sum = 0;
        vector<ll>a(n);
        for(int i=0;i<n;i++){
            cin>>a[i];
            sum+=a[i];
            maxi = max(maxi,a[i]);
        }

        ll sumk = sum+k;
        ll m = n;
        ll ans = 1;
        while(m>1){

            ll temp1 = sumk/m;
            if(m*temp1>=sum && temp1>=maxi){
                ans = m;
                break;
            }
            m--;

        }

        cout<<ans<<"\n";

    }
       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
1
5 10
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cards Partition
time_limit_ms: 2000
memory_limit_mb: 256
Description: [DJ Genki vs Gram - Einherjar Joker](https://soundcloud.com/leon- hwang-368077289/einherjar-joker-dj-genki-vs-gram)

You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy at most $k$ new cards in total, and the cards you buy can contain any integer between $\mathbf{1}$ and $\mathbf{n}$, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

  * all the decks must have the same size;    * there are no pairs of cards with the same value in the same deck.

Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^{16}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{10}$, $\sum a_i \geq 1$) — the number of cards of type $i$ you have at the beginning, for each $1 \leq i \leq n$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Examples:
input:
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
output:
2
3
1
7
2
2
1
1
2

Note: In the first test case, you can buy one card with the number $1$, and your cards become $[1, 1, 1, 1, 2, 2, 3, 3]$. You can partition them into the decks $[1, 2], [1, 2], [1, 3], [1, 3]$: they all have size $2$, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than $2$, so the answer is $2$.

In the second test case, you can buy two cards with the number $1$ and one card with the number $3$, and your cards become $[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5]$, which can be partitioned into $[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 5], [2, 3, 5], [2, 4, 5]$. You can show that you cannot get a partition with decks of size greater than $3$, so the answer is $3$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
LL a[N];
LL sum = 0, mxx = 1e18;
int n, k;
bool check(LL x) {
    LL c = sum / x;
    LL d =  (sum + k) / x;
    if (d >= mxx && (c < d || sum % x == 0)) {
        return true;
    }
    return false;
}
void solve() {
    cin >> n >> k;
    sum = 0, mxx = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        mxx = max(mxx, a[i]);
        sum += a[i];
    }
    for (LL i = n; i >= 1; i--) {
        if (check(i)) {
            cout << i << '\n';
            return;
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    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
1
3 1
3 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cards Partition
time_limit_ms: 2000
memory_limit_mb: 256
Description: [DJ Genki vs Gram - Einherjar Joker](https://soundcloud.com/leon- hwang-368077289/einherjar-joker-dj-genki-vs-gram)

You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy at most $k$ new cards in total, and the cards you buy can contain any integer between $\mathbf{1}$ and $\mathbf{n}$, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

  * all the decks must have the same size;    * there are no pairs of cards with the same value in the same deck.

Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^{16}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{10}$, $\sum a_i \geq 1$) — the number of cards of type $i$ you have at the beginning, for each $1 \leq i \leq n$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Examples:
input:
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
output:
2
3
1
7
2
2
1
1
2

Note: In the first test case, you can buy one card with the number $1$, and your cards become $[1, 1, 1, 1, 2, 2, 3, 3]$. You can partition them into the decks $[1, 2], [1, 2], [1, 3], [1, 3]$: they all have size $2$, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than $2$, so the answer is $2$.

In the second test case, you can buy two cards with the number $1$ and one card with the number $3$, and your cards become $[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5]$, which can be partitioned into $[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 5], [2, 3, 5], [2, 4, 5]$. You can show that you cannot get a partition with decks of size greater than $3$, so the answer is $3$.

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

public class AQ {
    static FastReader in = new FastReader();
    static final Random random = new Random();
    static long mod = 1000000007;
    static int MAX_N = (int) (1e5 + 1);
    static HashMap<Long, Integer> map = new HashMap<>();

    public static void main(String[] args) throws java.lang.Exception {
        StringBuilder res = new StringBuilder();
        int t = in.nextInt();

        while (t-- > 0) {
            int n = in.nextInt();
            long k = in.nextInt();
            int deckSize = n;
            long[] arr = in.readlongarray((int) n);

            long sum = Arrays.stream(arr).sum();
            long max = Arrays.stream(arr).max().orElseThrow();

            long maxCards = sum + k;

            while(deckSize > 1){
              long noOfDecks = maxCards / deckSize;

              if(noOfDecks * deckSize >= sum && noOfDecks >= max){
                res.append(deckSize).append("\n");
                break;
              }

              deckSize--;
            }

            if(deckSize == 1)
              res.append(1).append("\n");
        }

        print(res);
    }

    static long pow(long base, long num) {
        long res = 1;

        while (num > 0) {
            if (num % 2 == 1) {
                res = (res * base) % mod;
            }

            base = (base * base) % mod;
            num /= 2;
        }

        return res;
    }

    static int max(int a, int b) {
        if (a < b)
            return b;
        return a;
    }

    static int min(int a, int b) {
        if (a < b)
            return a;
        return b;
    }

    static void ruffleSort(int[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }

    static <E> void print(E res) {
        System.out.println(res);
    }

    static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    static int abs(int a) {
        if (a < 0)
            return -1 * a;
        return a;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int[] readintarray(int n) {
            int res[] = new int[n];
            for (int i = 0; i < n; i++)
                res[i] = nextInt();
            return res;
        }

        long[] readlongarray(int n) {
            long res[] = new long[n];
            for (int i = 0; i < n; i++)
                res[i] = nextLong();
            return res;
        }
    }

    static class SegmentTree {
        private int[] tree;
        private int[] lazy;
        private int size;

        public SegmentTree(int n) {
            size = n;
            tree = new int[size * 4];
            lazy = new int[size * 4];
        }

        public void build(int[] arr) {
            buildTree(arr, 0, 0, size - 1);
        }

        private void buildTree(int[] arr, int ind, int l, int r) {
            if (l == r) {
                tree[ind] = lazy[l];
                return;
            }

            int mid = (l + r) / 2;

            buildTree(arr, 2 * ind + 1, l, mid);
            buildTree(arr, 2 * ind + 2, mid + 1, r);

            tree[ind] = tree[2 * ind + 1] + tree[2 * ind + 2];
        }

        public int queryRangeSum(int l, int r) {
            return queryRangeUtilSum(l, r, 0, 0, size - 1);
        }

        private int queryRangeUtilSum(int l, int r, int ind, int start, int end) {
            if (lazy[ind] != 0) {
                tree[ind] = (r - l + 1) * lazy[ind];

                if (l != r) {
                    lazy[2 * ind + 1] += lazy[ind];
                    lazy[2 * ind + 2] += lazy[ind];
                }

                lazy[ind] = 0;
            }

            if (l > end || r < start || l > r) {
                return Integer.MIN_VALUE;
            }

            if (l >= start && r <= end) {
                return tree[ind];
            }

            int mid = (l + r) / 2;

            return queryRangeUtilSum(l, mid, 2 * ind + 1, start, end)
                    + queryRangeUtilSum(mid + 1, r, 2 * ind + 2, start, end);
        }

        public void rangeUpdate(int l, int r, int value) {
            rangeUpdateUtil(0, l, r, 0, size - 1, value);
        }

        private void rangeUpdateUtil(int ind, int l, int r, int start, int end, int value) {
            if (lazy[ind] != 0) {
                tree[ind] += (r - l + 1) * lazy[ind];

                if (r != l) {
                    lazy[2 * ind + 1] += lazy[ind];
                    lazy[2 * ind + 2] += lazy[ind];
                }

                lazy[ind] = 0;
            }

            if (l < start || r > end || l > r) {
                return;
            }

            if (start >= l && r <= end) {
                tree[ind] += (r - l + 1) * lazy[ind];

                if (r != l) {
                    lazy[2 * ind + 1] += lazy[ind];
                    lazy[2 * ind + 2] += lazy[ind];
                }

                return;
            }

            int mid = (l + r) / 2;
            rangeUpdateUtil(2 * ind + 1, l, mid, start, end, value);
            rangeUpdateUtil(2 * ind + 2, mid + 1, r, start, end, value);
        }
    }

    static int binarySearch(int[] arr, int key) {
        int left = 0;
        int right = arr.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (arr[mid] == key) {
                return mid;
            }

            if (arr[mid] < key) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return -1;
    }

    static long kadane(int[] arr) {
        long maxSum = Long.MIN_VALUE, currSum = 0;

        for (int value : arr) {
            currSum += value;
            if (currSum > maxSum) {
                maxSum = currSum;
            }
            if (currSum < 0) {
                currSum = 0;
            }
        }

        return maxSum;
    }
}

**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 code, we need to focus on edge cases where the algorithm might fail. The key points to consider are:

1. **Small `n` and `k` values**: These can help us understand how the algorithm handles small inputs.
2. **Large `n` and `k` values**: These can test the performance and correctness of the algorithm with large inputs.
3. **Zero values in `a_i`**: This can test how the algorithm handles cases where some card types are completely missing.
4. **High frequency of a single card type**: This can test how the algorithm handles cases where one card type dominates the others.
5. **Exact boundary conditions**: These can test how the algorithm handles edge cases where the number of cards and the number of coins are exactly enough to form a certain deck size.

Given the constraints and the need for a small-scale test input, let's construct a test case that combines these elements:

### Test Case
```plaintext
3
3 1
3 2 2
5 4
2 6 1 2 4
10 8
7 4 6 6 9 3 10 2 8 7
```

### Explanation
1. **First Test Case**:
   - `n = 3`, `k = 1`
   - `a = [3, 2, 2]`
   - This is a small input where the algorithm should be able to handle the addition of one card and partition the cards into decks of size 2.

2. **Second Test Case**:
   - `n = 5`, `k = 4`
   - `a = [2, 6, 1, 2, 4]`
   - This input has a higher number of cards and coins, and it tests the algorithm's ability to handle a more complex partitioning scenario.

3. **Third Test Case**:
   - `n = 10`, `k = 8`
   - `a = [7, 4, 6, 6, 9, 3, 10, 2, 8, 7]`
   - This input has a large number of card types and coins, and it tests the algorithm's ability to handle a large input and find the optimal deck size.

These test cases cover a range of scenarios and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cards Partition
time_limit_ms: 2000
memory_limit_mb: 256
Description: [DJ Genki vs Gram - Einherjar Joker](https://soundcloud.com/leon- hwang-368077289/einherjar-joker-dj-genki-vs-gram)

You have some cards. An integer between $1$ and $n$ is written on each card: specifically, for each $i$ from $1$ to $n$, you have $a_i$ cards which have the number $i$ written on them.

There is also a shop which contains unlimited cards of each type. You have $k$ coins, so you can buy at most $k$ new cards in total, and the cards you buy can contain any integer between $\mathbf{1}$ and $\mathbf{n}$, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

  * all the decks must have the same size;    * there are no pairs of cards with the same value in the same deck.

Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $k$ ($1 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^{16}$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 10^{10}$, $\sum a_i \geq 1$) — the number of cards of type $i$ you have at the beginning, for each $1 \leq i \leq n$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Examples:
input:
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
output:
2
3
1
7
2
2
1
1
2

Note: In the first test case, you can buy one card with the number $1$, and your cards become $[1, 1, 1, 1, 2, 2, 3, 3]$. You can partition them into the decks $[1, 2], [1, 2], [1, 3], [1, 3]$: they all have size $2$, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than $2$, so the answer is $2$.

In the second test case, you can buy two cards with the number $1$ and one card with the number $3$, and your cards become $[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5]$, which can be partitioned into $[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 5], [2, 3, 5], [2, 4, 5]$. You can show that you cannot get a partition with decks of size greater than $3$, so the answer is $3$.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class C2019 {

	private static Scanner s = new Scanner(System.in);

	public static void main(String args[]) {
		int t = s.nextInt();
		while (t-->0) solve();
	}

	public static void solve() {
		long n = s.nextInt(), k = s.nextInt();
		ArrayList<Long> A = new ArrayList<>();

		for (int i=0;i<n;i++) A.add(s.nextLong());

		A.sort(null);
		long sum = 0;
		for (int i=0;i<n;i++) sum+=A.get(i);

		long mx = A.get((int) (n-1));

		for (long i=n;i>0;i--) {
			long x = i*mx;
			x = Math.max(x, sum + (i - sum%i)%i);

			if (x-sum<=k) {
				System.out.println(i);
				return;
			}
		}
	}

}

**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
3 2
2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
// LUOGU_RID: 158207122
#include<iostream>
#include<cstring>
using namespace std;
bool chayi(string a,string b) {
	int count=0;
	for(int i=0; i<a.length(); i++) {
		if(a[i]!=b[i]) {
			count++;
		}
		if(count>1) {
			return 1;//差异字符量大于1时返回1
		}
	}
	return 0;
}
int main() {
	string s,scut;
	int t,n;
	cin>>t;
	for(int i=0; i<t; i++) {
		cin>>n;
		cin>>s;
		int flag=0;
		if(s.length()==1) {
			cout<<1<<endl;
		}
		bool a[200000]={0};
		for(int j=1; j<=n/2; j++) {
			if(n%j!=0);
			for(int k=0; k<n; k+=j) {
				if(a[k]==1)
				{
					continue;
				}
				scut=s.substr(k,j);
				flag=0;
				for(int kk=0; kk<n; kk+=j) {
					string s2=s.substr(kk,j);
					if(scut==s2)
					{
						a[kk]=1;
					}
					else if(flag==1) {
						flag=2;
						break;
					} else {
						if(chayi(scut,s2)) {
							flag=2;
							break;
						} else {
							flag=1;
						}
					}
				}
				if(flag!=2) {
					cout<<j<<endl;
					break;
				}
			}
			if(flag!=2) {
				break;
			}

		}
		if(flag==2) {
			cout<<n<<endl;
		}
	}
}

**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
4
abab
6
abcabc
8
abababab
12
abcabcabcabc
15
abcdeabcdeabcde
```

5
4
abab
6
abcabc
8
abababab
12
abcabcabcabc
15
abcdeabcdeabcde


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n,len,max_=0,min_=10000000;
		int v[300]={0};
		string s;
		cin>>n>>s;
		for(int i=0;i<n;i++)
		{
			v[int(s[i])]++;
		}
		for(int i=0;i<n;i++)
		{
			max_=max(max_,v[int(s[i])]);
			min_=min(min_,v[int(s[i])]);
		}
		if(max_+min_==n)
		{
			if(min_==1)
			{
				cout<<"1\n";
				continue;
			}

		}
		if(min_>=max_-2)len=n/((max_+min_)/2);
		else
		{
			len=n/max_;//cout<<len;
		}
		//cout<<len;
		string s1="",s2="";
		for(int i=0;i<len;i++)
		{
			s1+=s[i];
			s2+=s[n-len+i];
		}
		int num1=0,num2=0,cnt=0;
		for(int i=0;i<n;i++)
		{
			if(num1>=2||num2>=2)
			{
				cout<<n<<"\n";
				cnt=1;
				break;
			}
			if(i%len==0)num1=num2=0;
			if(s1[i%len]!=s[i])
			{
				num1++;
			}
			if(s2[i%len]!=s[i])
			{
				num2++;
			}
		}
		if(!cnt)
		{
			cout<<len<<"\n";
		}
	}

	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
1
12
aabaabaaabaa
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <climits>
#include <cmath>
#include <iomanip>

using namespace std;

ifstream fin("data.in");
ofstream fout("data.out");

#define fin cin
#define fout cout

const double margin = 0.000001;

int cmpD(double x, double y);
void solveTest();
bool checkStrD(string& str, long long d);

int main()
{
    int tests;

    fin >> tests;

    while(tests--)
        solveTest();

    return 0;
}

int cmpD(double x, double y)
{
    if(abs(x - y) < margin)
        return 0;

    if(x - y < margin)
        return -1;

    return 1;
}
void solveTest()
{
    long long n, best = LLONG_MAX;
    string str;

    fin >> n >> str;

    for(long long d = 1; d * d <= n; d++)
    {
        if(checkStrD(str, d))
        {
            best = min(best, d);
        }
        if(checkStrD(str, n / d))
        {
            best = min(best, n / d);
        }
    }

    fout << best << '\n';
}
bool checkStrD(string& str, long long d)
{
    int errors = 0, start = 0;

    if(d == str.size())
        return true;

    for(int i = 0, j = str.size() - d; i < d; i++, j++)
    {
        if(str[i] != str[j])
        {
            errors++;
        }
    }

    if(errors > 2)
        return false;
    else if(errors == 1)
    {
        start = d;
    }

    errors = 0;

    for(int i = 0; i < str.size(); i += d)
    {
        for(int j = i, st = start; st < start + d; j++, st++)
        {
            if(str[st] != str[j])
            {
                errors++;

                if(errors > 1)
                    return false;
            }
        }
    }

    return true;
}

**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
12
abababababab
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
def func(s1,s2):

    count=0
    for i in range(len(s1)):
        if s1[i]!=s2[i]:
            count+=1
    return count

t = int(input())
for _ in range(t):
    n = int(input())
    s = input()
    ans=""
    mini = float('inf')
    for i in range(len(s)):
        for j in range(i,len(s)):
            if(n%(j-i+1)==0):
                ans = s[i:j+1]*(n//(j-i+1))
                if(func(ans,s)==0 or func(ans,s)==1):
                    if(func(ans,s)<mini):
                        min_ans = j-i+1

    print(min_ans)

**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 and scenarios where the code might fail. The main issues in the buggy code are:

1. **Inefficient Comparison**: The code checks every possible substring, which can be very slow for large strings.
2. **Incorrect Handling of Substrings**: The code does not correctly handle cases where the substring needs to be repeated multiple times to match the length of the original string.
3. **Edge Cases**: The code might fail for specific edge cases, such as strings with very few unique characters or strings that are already repeating.

Let's consider the following scenarios to create a challenging test input:

1. **Strings with Repeated Patterns**: These can be tricky because the code might not correctly identify the shortest repeating substring.
2. **Strings with a Single Character Difference**: The code should be able to handle cases where the string is almost a perfect repetition but has one character different.
3. **Strings with Minimal Unique Characters**: These can be challenging because the code needs to correctly identify the shortest repeating pattern.

Given these considerations, here is a small-scale test input that can expose the bug:

```plaintext
3
6
aaaaaa
6
aaaaab
6
ababab
```

Explanation:
1. **Test Case 1**: `aaaaaa`
   - The shortest repeating substring is `a`, and it repeats 6 times. The code should output `1`.
2. **Test Case 2**: `aaaaab`
   - The shortest repeating substring is `a`, and it repeats 5 times, with one character different. The code should output `1`.
3. **Test Case 3**: `ababab`
   - The shortest repeating substring is `ab`, and it repeats 3 times. The code should output `2`.

This input is designed to test the code's ability to handle repeated patterns and single character differences, which are common edge cases that can expose bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
import sys
from math import ceil, inf
from collections import defaultdict , Counter
mod = pow(10, 9) + 7

def ip():
  return int(input())

def input():
  return sys.stdin.readline().strip()

def readList():
  return list(map(int,input().split()))

def solve():
  n = ip()
  s = input()
  if n % 2:
    print(n)
    return
  counter = Counter()
  right = 0

  while right < n:
    counter[s[right]] += 1
    if counter[s[right]] > 1 and n % (right) == 0:
      # print(right)
      break
    right += 1

  k = s[:right] * (n // (right - 1))
  count = set()

  # print(k)
  for i in range(n):
    if s[i] != k[i]:
      count.add(k[i])

  option2 = s[0] * n
  count2 = []
  for i in range(n):
    if s[i] != option2[i]:
      count2.append(s[i])

  print(1 if len(count2) < 2 else right if len(count) < 2 else n)

T = int(input())
# T = 1

for _ in range(T):
  solve()

**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 consider edge cases and scenarios where the code might fail. Here are some key points to consider:

1. **Odd Length Strings**: The code immediately returns the full length `n` if `n` is odd. This is a valid scenario, but we should ensure it's handled correctly.
2. **Repeating Patterns**: The code tries to find repeating patterns and breaks if a repeating character is found. We need to create a string where this logic might fail.
3. **Single Character Repeating**: The code also checks if a single character can repeat to form the string. We need to ensure that this is handled correctly.
4. **Edge Cases with Small Lengths**: Small lengths can sometimes expose bugs in the logic.

Let's create a test input that includes a string where the repeating pattern is not straightforward and the code might fail to find the correct shortest repeating substring.

```plaintext
3
10
abcabcabca
5
ababa
7
abacaba
```

### Explanation:
1. **Test Case 1**:
   - **Input**: `10 abcabcabca`
   - **Expected Output**: `3`
   - **Explanation**: The string "abcabcabca" can be formed by repeating "abc" three times and then adding "a". The shortest repeating substring is "abc", which has a length of 3.

2. **Test Case 2**:
   - **Input**: `5 ababa`
   - **Expected Output**: `5`
   - **Explanation**: The string "ababa" is a palindrome and does not have a shorter repeating substring. The shortest repeating substring is "ababa", which has a length of 5.

3. **Test Case 3**:
   - **Input**: `7 abacaba`
   - **Expected Output**: `7`
   - **Explanation**: The string "abacaba" does not have a shorter repeating substring. The shortest repeating substring is "abacaba", which has a length of 7.

These test cases cover different scenarios and should help expose any issues in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
def check(size, s):
    if len(s) % size != 0: return False
    mul = len(s) // size

    for i in range(N - size + 1):
        diff = 0
        ss = s[i:i+size] * mul
        for i in range(len(s)):
            diff += s[i] != ss[i]
        if diff < 2: return True

    return False

for _  in range(int(input())):
    n = int(input())
    s = input()
    res = N = len(s)

    left = 1
    right = N
    while left <= right:
        mid = (left + right) // 2
        if check(mid, s):
            res = mid
            right = mid - 1
        else:
            left = mid + 1

    print(res)

# 1
# for _  in range(int(input())):
# a,b,c = map(int, input().split())

# if a < b and b < c: print('STAIR')
# elif a < b and b > c: print('PEAK')
# else: print('NONE')
# ---------------------
# ---------------------

# 2.
# for _  in range(int(input())):
#     n = int(input())

#     for row in range(n):
#         if row % 2 == 0:
#             r = ''
#             for col in range(n):
#                 if col % 2 == 0: r += '##'
#                 else: r += '..'
#             print(r)
#             print(r)
#         else:
#             r = ''
#             for col in range(n):
#                 if col % 2 == 0: r += '..'
#                 else: r += '##'
#             print(r)
#             print(r)

# 3
# for _  in range(int(input())):
#     hour, minute = map(int,input().split(':'))

#     if hour == 0:
#         print('12:' + '{:02d}'.format(minute) + ' AM')
#     elif hour < 12:
#         print('{:02d}:{:02d}'.format(hour, minute) + ' AM')
#     elif hour == 12:
#         print('12:' + '{:02d}'.format(minute) + ' PM')
#     else:
#         print('{:02d}:{:02d}'.format(hour - 12, minute) + ' PM')

**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 focus on edge cases that can cause the algorithm to fail. The primary issues to consider are:

1. **Small lengths of the string** where the repeating pattern is very short.
2. **Strings with a single character** where the repeating pattern might be the entire string.
3. **Strings with a specific repeating pattern** that might be close to the correct answer but not quite.
4. **Strings with a single differing character** that should be considered in the nearly repeating pattern.

Given these considerations, let's construct a test input that can highlight these issues.

### Test Input
```plaintext
5
4
aaaa
4
abba
6
abcabc
7
aabaaba
10
abcabcabca
```

### Explanation
1. **"aaaa"** (length 4):
   - The shortest repeating substring is "a" (length 1), which matches the string exactly.
2. **"abba"** (length 4):
   - The shortest repeating substring is "abba" (length 4), as there is no shorter substring that can form "abba" with at most one differing character.
3. **"abcabc"** (length 6):
   - The shortest repeating substring is "abc" (length 3), which matches the string exactly.
4. **"aabaaba"** (length 7):
   - The shortest repeating substring is "aabaaba" (length 7), as there is no shorter substring that can form "aabaaba" with at most one differing character.
5. **"abcabcabca"** (length 10):
   - The shortest repeating substring is "abcabc" (length 6), but the correct answer should be "abc" (length 3) because "abcabcabca" can be formed by repeating "abc" and changing the last character.

This test input should help expose the bug in the code by covering various edge cases and scenarios where the algorithm might fail.

5
4
aaaa
4
abba
6
abcabc
7
aabaaba
10
abcabcabca


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
import java.util.*;
public class Main
{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while(t-- >0)
        {
            int n = scanner.nextInt();
            String s = scanner.next();
            if(n%2==1)
            {
                System.out.println(n);
                continue;
            }
            int ans = n;
            for(int k=n/2;k > 1;k/=2)
            {
                int c = 0;
                for(int i=0,j=k;i<k;i++,j++) {
                    if(s.charAt(i)!=s.charAt(j))++c;
                }
                if(c > 1)
                    break;
                ans = k;
            }
            if(ans == 2) {
                int[] ch = new int[26];
                for (char c : s.toCharArray())
                    ++ch[c - 'a'];

                int v1 = 0, v2 = 0,tc=0;
                for (int val : ch)
                {
                    if(val > 0)
                    {
                        ++tc;
                        if(v1 == 0)
                            v1 = val;
                        else
                            v2 = val;
                    }
                }
                if(tc == 2 && (v1 == 1 && v2 == n-1) || (v2 == 1 && v1 == n-1))
                    ans = 1;
            }
            System.out.println(ans);
        }

    }
}

**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
6
abcabc
8
aabbccdd
10
abcdeabcde
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
    public static Read scanner = new Read(System.in);
    //    public static final long MOD = (long) (1e9 + 7);
    public static final long MOD = 998244353;
    public static final int MID = (int) (1 * 1e2 + 7);
    public static final long LMAX = Long.MAX_VALUE;
    public static final long LMIN = Long.MIN_VALUE;
    public static final int IMAX = Integer.MAX_VALUE;
    public static final int IMIN = Integer.MIN_VALUE;
    public static boolean pool = false;
    public static int[] arr = new int[MID];
//    public static int[] t = new int[MID];
//    public static int[] c = new int[MID];
//    public static long[] Q = new long[MID];
//    public static long[] C = new long[MID];
//    public static long[] H = new long[MID];
    public static int N, M, n, m , k;
    public static long ans = LMAX;
    public static long num;
    public static long cnt = 0;
//    public static boolean[] stack = new boolean[MID];
    //四个方向
    public static int[] row = {1, -1, 0, 0};
    public static int[] cos = {0, 0, 1, -1};
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    public static void solve() {
        int n = scanner.nextInt();
        String str = scanner.nextLine();
        str = " " + str;
        for(int i = 1 ; i <= n ; i++){
            if(n % i == 0){
                String s = str.substring(1 , 1 + i);
                for(int k = 1 ; k <= n ; k += i){
                    s = str.substring(k , k + i);
                    cnt = 0;
                    for(int j = 1 ; j <= n ; j += i){
                        int o = 0;
                        for(int index = j ; index < j + i ; index++){
                            if(s.charAt(o++) != str.charAt(index)) cnt++;
                        }
                        if(cnt > 1) break;
                    }
                }

            }
            if(cnt <= 1){
                ans = i;
                break;
            }
        }
        System.out.println(ans);

    }
    public static void main(String[] args) {

//        long start = System.currentTimeMillis();//程序开始时间戳
        int T = scanner.nextInt();
        while (T-- != 0) {
            solve();
        }
        out.flush();
//        long end = System.currentTimeMillis();//程序结束时间戳
//        out.println("运行时间 : " + (end - start));//输出运行时间
    }

    //快读
    static class Read {
        BufferedReader br;
        StringTokenizer st;

        public Read(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in), 16384);
            eat("");
        }

        public void eat(String s) {
            st = new StringTokenizer(s);
        }

        public String nextLine() {
            try {
                return br.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public boolean hasNext() {
            while (!st.hasMoreTokens()) {
                String s = nextLine();
                if (s == null)
                    return false;
                eat(s);
            }
            return true;
        }

        public String next() {
            hasNext();
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public Double nextDouble() {
            return Double.parseDouble(next());
        }

        public BigInteger nextBigInteger() {
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal() {
            return new BigDecimal(next());
        }
    }

    public static void swap(long[] arr, int index1, int index2) {
        long t = arr[index1];
        arr[index1] = arr[index2];
        arr[index2] = t;
    }

    public static long Dot_pitch(long x, long y, long xx, long yy) {
        return (xx - x) * (xx - x) + (yy - y) * (yy - y);
//        return (long) Math.sqrt(Math.pow(xx - x , 2) + Math.pow(yy - y , 2));
    }
}

class way {
    public long h , a;

    public way(long h, long a) {
        this.h = h;
        this.a = a;
    }
}

**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 code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. The code's main issue lies in how it checks the repeating substrings and the conditions under which it counts mismatches.

### Steps to Generate the Test Input:
1. **Edge Case: Single Character String**:
   - A string of length 1 should always return 1, as it is the shortest repeating substring.
   - Example: `s = "a"`

2. **Edge Case: All Characters Same**:
   - A string where all characters are the same should return 1, as the shortest repeating substring is the single character.
   - Example: `s = "aaaaa"`

3. **Edge Case: Two Different Characters**:
   - A string with only two different characters can be tricky, especially if the length is not a multiple of 2.
   - Example: `s = "ababab"`

4. **Edge Case: Nearly Repeating Substring**:
   - A string that is nearly repeating but has one character different.
   - Example: `s = "abcabcabc"`

5. **Edge Case: Length Not a Multiple of Substring Length**:
   - A string where the length is not a multiple of the repeating substring length.
   - Example: `s = "abcabcabc"`

6. **Edge Case: Large String**:
   - A large string that is nearly repeating but has one character different, to test performance and correctness.
   - Example: `s = "abcabcabcabcabcabcabcabcabcabc"`

### Challenging Test Input:
```plaintext
5
1
a
5
aaaaa
6
ababab
9
abcabcabc
20
abcabcabcabcabcabcabcabcabcabc
```

### Explanation:
- **Test Case 1**: `s = "a"` - The shortest repeating substring is "a" with length 1.
- **Test Case 2**: `s = "aaaaa"` - The shortest repeating substring is "a" with length 1.
- **Test Case 3**: `s = "ababab"` - The shortest repeating substring is "ab" with length 2.
- **Test Case 4**: `s = "abcabcabc"` - The shortest repeating substring is "abc" with length 3.
- **Test Case 5**: `s = "abcabcabcabcabcabcabcabcabcabc"` - The shortest repeating substring is "abc" with length 3.

This test input covers various edge cases and should help expose any issues in the buggy code.

5
1
a
5
aaaaa
6
ababab
9
abcabcabc
20
abcabcabcabcabcabcabcabcabcabc


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
/*

Author: Kolyanchick
Date: 28-03-2024

 |  Kolyanchick  |
\|/             \|/

⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠃
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠛⠛⠋⠉⠉⠉⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡏
⠀⢹⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣤⣤⣴⣶⣶⣶⣾⣾⣿⣿⣟⣻⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡀
⣶⣾⣿⣿⣿⣿⣿⣿⠿⠟⠋⣓⣻⠿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣽⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡆
⣿⣿⣿⣿⣿⣿⡿⢃⣦⡞⣿⣫⢧⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠛⠿⠿
⣿⣿⣿⡿⡿⠋⣴⣿⡿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣅⢻⡿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡄
⣿⣟⡝⠀⣠⣾⡿⢫⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣭⡻⣿⣿⣿⣞⣷⣍⣿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⡄
⢋⣼⢤⣼⠋⣲⣵⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡗⠛⣻⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠇
⣾⢓⡿⢳⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⢿⣿⣿⣿⣿⣿⣿⣿⡿⠏⠁⠙⠿⢿⡋⠡⣾⣟⣹⣿⣿⣿⣿⣿⣿⣿⣿⣿⠁
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣏⠁⠀⠀⠀⠀⠀⠉⠉⠉⠙⠋⠀⠀⠀⠀⠀⠀⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣤⣤⣄
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⡄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣇
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠉⡉⠐⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⣶⣶⣶⡀⢤⠀⢀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠈⠻⢿⣷⡘⣷⣄⠀⡀⢀⠀⣘⢀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣤⣶⣶⣾⣿⣦⣀⣙⣀⡿⣸⠓⢻⢸⣧⣴⣤⡀⠀⠐⠤⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡏⠀⣹
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠇⢸⣿⣿⣿⣿⣿⣿⣗⡀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠟⠛⠿⠿⠿⠿⠿⠿⠟⢻⣿⣿⣿⣏⠀⠀⢟⣿⡯⠥⠤⠉⠉⠀⠘⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠡⠀⠀⠀⠀⠀⠀⠀⠘⢩⣿⡿⣿⣿⣆⠀⠀⠉⠁⠀⠀⠀⠀⠀⠀⣏⠀⠀⠀⠀⠈⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣄⠀⢠⠀⠀⠀⠀⠀⠀⠈⣿⣧⣿⣿⡿⠀⠀⣀⠀⠀⠀⠀⠀⠀⠀⡏⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣤⡀⠀⠀⠀⠀⢠⣿⡿⣿⣿⡇⠀⠀⠘⣷⡀⠀⠀⠀⠀⢰⡇⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⣄⠀⢠⣿⣿⣿⣿⣿⣇⠀⢀⠄⠹⣷⠀⠀⠀⠀⣸⣷⣄⣀⣠⣤⣤⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠀⠈⢿⣿⣿⣿⣿⣿⠿⠋⠀⠀⠉⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⢰⡀⠈⠛⠛⣛⠉⠁⠀⠀⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⢠⣴⣤⣭⣴⣶⣴⡶⠶⢦⡄⠀⠀⢠⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣴⡉⠛⢙⢡⡴⠋⠀⠀⢀⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣿⡿⠋⡀⠀⠀⠀⣼⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⡏⡞⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠛⠿⣷⡿⠋⠽⠟⠁⠀⢾⠀⠀⡞⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⡿⣿⠁⣧⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣄⠀⠈⠃⠀⠀⠀⠀⠀⠀⣀⣾⠇⠸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⠟⢱⣿⡄⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⣤⣤⣤⣠⣤⣄⣤⣶⣿⡟⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⠟⠃⣀⣸⣿⣷⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⣿⣿⣿⣿⡞⠀⠂⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣶⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠋⠁⠀⠀⠀⣴⠏⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠟⠋⣀⡀⠀⠀⣀⣴⠟⠁⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠛⠉⠁⠀⠀⠈⠙⢤⡀⠀⠀⣠⣴⠟⠁⠀⠀⠀⠀⠀⠀⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡀⠀⠀⠀⠀⠀⠀⠀⣈⣓⣚⣿⣁⣀⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣇⣀⣀⣀⣀⣤⣶⣿⣿⣿⣯⣏⣟⣋⣿⣶⣦⣄⣀⣀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿

*/

#include <bits/stdc++.h>
#define int long long
#define pb push_back

using namespace std;

const int N = 1e5 + 10;
bool f[N];

void sol() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<int> e;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            e.pb(i);
            if (i * i != n) e.pb(n / i);
        }
    }
    sort(e.begin(), e.end());
    for (int x : e) {
        int c = 0, now = 0;;
        for (int i = 0; i < n - x; i++) {
            if (s[i] != s[i + x] && c == 0) {
                c++;
                now = i;
            } else if (s[i] != s[i + x]) {
                if (i == now + x) continue;
                c++;
                break;
            }
        }
        if (c <= 1) {
            cout << x << '\n';
            return;
        }
    }
    return;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;
    cin >> t;
    while (t--) sol();
    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
5
4
abab
6
abcabc
8
abcdabcd
10
abcdeabcde
12
abcdeabcdeab
```

5
4
abab
6
abcabc
8
abcdabcd
10
abcdeabcde
12
abcdeabcdeab


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

#define intll long long
#define endl '\n'
#define TEST_CASE true

template<typename T>
void PRINT(vector<T>& arr) {
	for (int i = 0; i < int(arr.size()); i++)
		cout << arr[i] << " \n"[i == int(arr.size()) - 1];
}

int mod1 = 127657753, mod2 = 987654319;
int expo(long long x, long long y, int mod) {
	int ans = 1; x %= mod;
	while (y) {
		if (y & 1) ans = 1ll * ans * x % mod;
		x = 1ll * x * x % mod; y = y >> 1;
	}
	return (ans + mod) % mod;
}

using HASH = pair<int, int>;
HASH add(HASH a, HASH b) {
	HASH res;
	res.first = ((long long) a.first + b.first) % mod1;
	res.second = ((long long) a.second + b.second) % mod2;
	return res;
}

HASH subs(HASH a, HASH b) {
	HASH res;
	res.first = (a.first - b.first) % mod1;
	res.second = (a.second - b.second) % mod2;
	res.first = (res.first + mod1) % mod1;
	res.second = (res.second + mod2) % mod2;
	return res;
}

HASH mult(HASH a, HASH b) {
	HASH res;
	res.first = (1ll * a.first * b.first) % mod1;
	res.second = (1ll * a.second * b.second) % mod2;
	return res;
}

const int sizee = 2e5 + 10;
HASH pw{137, 277};
HASH invp{0, 0};
vector<HASH> power(sizee), invpower(sizee);

void calculation() {
	power[0] = {1, 1};
	for (int i = 1; i < sizee; i++) {
		power[i] = mult(power[i - 1], pw);
	}
	invp.first = expo(pw.first, mod1 - 2, mod1);
	invp.second = expo(pw.second, mod2 - 2, mod2);
	invpower[0] = {1, 1};
	for (int i = 1; i < sizee; i++) {
		invpower[i] = mult(invpower[i - 1], invp);
	}
}

struct hashing {
	int N;
	vector<HASH> hash; // 1 - indexed
	hashing() {}
	hashing(string s) { build(s); }

	void build(string &s) {
		N = int(s.size());
		hash.push_back({0, 0});
		for (int i = 0; i < N; i++) {
			HASH P = add(hash[i], mult(power[i], {s[i], s[i]}));
			hash.push_back(P);
		}
	}

	// hash = (h[r] - h[l - 1]) / p[l - 1]
	HASH query(int l, int r) { // 1 - indexed
		if (l > r) return {0, 0};
		if (l < 1 || r > N) return {0, 0};
		assert(1 <= l && l <= r && r <= N);
		HASH ans = mult(subs(hash[r], hash[l - 1]), invpower[l - 1]);
		return ans;
	}
};

struct Mint {
	uint64_t n;
	static uint64_t mod, inv, r2;
	Mint() : n(0) { }
	Mint(const uint64_t &x) : n(init(x)) { }
	static uint64_t init(const uint64_t &w) { return reduce(__uint128_t(w) * r2); }
	static void set_mod(const uint64_t &m) {
		mod = inv = m;
		for (int i = 0; i < 5; i++) {
			inv *= 2 - inv * m;
		}
		r2 = -__uint128_t(m) % m;
	}
	static uint64_t reduce(const __uint128_t &x) {
		uint64_t y = uint64_t(x >> 64) - uint64_t((__uint128_t(uint64_t(x) * inv) * mod) >> 64);
		return int64_t(y) < 0 ? y + mod : y;
	}
	Mint& operator+= (const Mint &x) { n += x.n - mod; if (int64_t(n) < 0) n += mod; return *this; }
	Mint& operator+ (const Mint &x) const { return Mint(*this) += x; }
	Mint& operator*= (const Mint &x) { n = reduce(__uint128_t(n) * x.n); return *this; }
	Mint& operator* (const Mint &x) const { return Mint(*this) *= x; }
	uint64_t val() const { return reduce(n); }
};

uint64_t Mint::mod, Mint::inv, Mint::r2;

bool suspect(const uint64_t &a, const uint64_t &s, uint64_t d, const uint64_t &n) {
	if (Mint::mod != n)  Mint::set_mod(n);
	Mint x(1), xx(a), one(x), minusone(n - 1);
	while (d > 0) {
		if (d & 1)
			x *= xx;
		xx *= xx;
		d >>= 1;
	}
	if (x.n == one.n)
		return true;
	for (unsigned int r = 0; r < s; r++) {
		if (x.n == minusone.n)
			return true;
		x *= x;
	}
	return false;
}

bool is_prime(const uint64_t &n) {
	if (n <= 1 || (n > 2 && (~n & 1)))
		return false;
	uint64_t d = n - 1, s = 0;
	while (~d & 1)
		s++,
		d >>= 1;
	static const uint64_t a_big[] = {
		2,
		325,
		9375,
		28178,
		450775,
		9780504,
		1795265022
	};
	static const uint64_t a_smol[] = {
		2,
		7,
		61
	};
	if (n <= 4759123141LL) {
		for (auto && p : a_smol) {
			if (p >= n)
				break;
			if (!suspect(p, s, d, n))
				return false;
		}
	}
	else {
		for (auto && p : a_big) {
			if (p >= n)
				break;
			if (!suspect(p, s, d, n))
				return false;
		}
	}
	return true;
}

uint64_t rho_pollard(const uint64_t &n) {
	if (~n & 1)
		return 2u;
	static random_device rng;
	uniform_int_distribution<uint64_t> rr(1, n - 1);
	if (Mint::mod != n)
		Mint::set_mod(n);
	for (;;) {
		uint64_t c_ = rr(rng), g = 1, r = 1, m = 500;
		Mint y(rr(rng)), xx(0), c(c_), ys(0), q(1);
		while (g == 1) {
			xx.n = y.n;
			for (unsigned int i = 1; i <= r; i++)
				y *= y,
				     y += c;
			uint64_t k = 0; g = 1;
			while (k < r && g == 1) {
				for (unsigned int i = 1; i <= (m > (r - k) ? (r - k) : m); i++) {
					ys.n = y.n;
					y *= y; y += c;
					uint64_t xxx = xx.val(), yyy = y.val();
					q *= Mint(xxx > yyy ? xxx - yyy : yyy - xxx);
				}
				g = __gcd<uint64_t>(q.val(), n);
				k += m;
			}
			r *= 2;
		}
		if (g == n)
			g = 1;
		while (g == 1) {
			ys *= ys; ys += c;
			uint64_t xxx = xx.val(), yyy = ys.val();
			g = __gcd<uint64_t>(xxx > yyy ? xxx - yyy : yyy - xxx, n);
		}
		if (g != n && is_prime(g))
			return g;
	}
	assert(69 == 420);
}

template <typename T>
vector<T> factor(T n) {
	if (n < 2)
		return { };
	if (is_prime(n))
		return {n};
	T d = rho_pollard(static_cast<uint64_t>(n));
	vector<T> left = factor<T>(d);
	vector<T> right = factor<T>(n / d);
	left.insert(left.end(), right.begin(), right.end());
	return left;
}

template<typename T = int>
vector<pair<T, int>> factorize(T n) {
	auto primes_fact = factor<T>(n);
	sort(primes_fact.begin(), primes_fact.end());
	if (primes_fact.empty())
		return {};
	vector<pair<T, int>> fact;
	T p = primes_fact[0];
	int cnt = 1;
	for (int i = 1; i < (int)primes_fact.size(); i++) {
		if (primes_fact[i] == p) cnt++;
		else {
			fact.emplace_back(p, cnt);
			p = primes_fact[i], cnt = 1;
		}
	}
	fact.emplace_back(p, cnt);
	return fact;
}

template<typename T = int>
int nod(vector<pair<T, int>>& fact) {
	int ans = 1;
	for (auto i : fact) ans *= 1ll * (i.second + 1);
	return ans;
}

template<typename T = int>
int nod(T n) {
	auto fact = factorize<T>(n);
	return nod(fact);
}

template<typename T = int>
T sod(vector<pair<T, int>>& fact) {
	T ans = 1;
	for (auto i : fact) {
		T p = i.first;
		T total = 1;
		int m = i.second;
		while (m--) {
			total += p;
			p *= 1ll * i.first;
		}
		ans *= 1ll * total;
	}
	return ans;
}

template<typename T = int>
T sod(T n) {
	auto fact = factorize<T>(n);
	return sod(fact);
}

template<typename T = int>
T phi(vector<pair<T, int>>& fact, T n) {
	if (n == 1) return 1;
	T ans = n;
	for (auto i : fact) ans -= (ans / i.first);
	return ans;
}

template<typename T = int>
T phi(T n) {
	auto fact = factorize<T>(n);
	return phi(fact, n);
}

template<typename T = int>
void generate_divisor(int idx, T div, vector<pair<T, int>>&fact, vector<T>& divisor) {
	if (idx == int(fact.size())) {
		divisor.emplace_back(div);
		return;
	}
	for (int i = 0; i <= fact[idx].second; i++) {
		generate_divisor(idx + 1, div, fact, divisor);
		div = 1ll * div * fact[idx].first;
	}
}

template<typename T = int>
vector<T> get_divisor(vector<pair<T, int>>&fact) {
	vector<T> divisor;
	generate_divisor((T)0, (T)1, fact, divisor);
	sort(divisor.begin(), divisor.end());
	return divisor;
}

// O(sqrt(n))
template<typename T = int>
vector<T> get_divisor(T n) {
	auto fact = factorize<T>(n);
	return get_divisor(fact);
}

// 2 to n primes
template<typename T = int>
vector<T> get_prime(T n) {
	vector<T> prime{2};
	for (int i = 3; i <= n; i += 2) {
		if (is_prime(i))
			prime.emplace_back(i);
	}
	return prime;
}

void solution() {

	int n;
	cin >> n;

	string s;
	cin >> s;

	hashing H(s);
	auto div = get_divisor(n);

	for (auto i : div) {
		map<HASH, int> cnt;
		map<HASH, HASH> interval;
		for (int j = 1; j <= n; j += i) {
			auto H1 = H.query(j, j + i - 1);
			cnt[H1]++;
			interval[H1] = {j, j + i - 1};
		}

		if (cnt.size() > 2) continue;
		auto it = interval.begin();
		if (cnt.size() == 2) {
			for (auto [k, c] : cnt) {
				if (c == 1) {
					auto [l1, r1] = it->second;
					++it;
					auto [l2, r2] = it -> second;
					int same = 0;
					for (int k = 0; k < i; k++) {
						same += (s[l1 - 1] != s[l2 - 1]);
						l1++;
						l2++;
						if (same >= 2) break;
					}
					if (same == 1) {
						return void(cout << i << endl);
					}
					break;
				}
			}
		}
	}

	cout << n << endl;
}

int32_t main() {

	ios::sync_with_stdio(false) ; cin.tie(0) ;

	calculation();

	int t_c = 1, tt_c = 1;
	if (TEST_CASE) cin >> t_c;

	while (t_c--) {
		// cout << "Case " << tt_c++ << ": ";
		solution();
		// if (t_c) cout << '\n';
	}

	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 generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios where the code might fail. Here are the steps to think through this:

1. **Understand the Problem**:
   - We need to find the shortest string \( k \) such that several copies of \( k \) can be concatenated to form a string with the same length as \( s \) and, at most, one different character.

2. **Identify Potential Bugs**:
   - The code uses hashing and prime factorization to find the divisors of the length of the string \( s \).
   - It then checks if the string can be divided into repeating segments with at most one different character.
   - Potential issues could arise from:
     - Incorrect handling of edge cases (e.g., very short strings, strings with all identical characters).
     - Incorrect comparison of hash values.
     - Incorrect calculation of the number of differing characters.

3. **Generate Test Input**:
   - We need a test input that is small but challenging enough to expose these issues.
   - A string with a length that has multiple divisors and contains a pattern with one differing character is a good candidate.

4. **Example Test Input**:
   - Consider a string of length 12, which has divisors 1, 2, 3, 4, 6, and 12.
   - The string can be "abababababab" with one character changed to "c" to make it "abababacabab".

Here is the test input:

```plaintext
1
12
abababacabab
```

This input will challenge the code to correctly identify the shortest repeating substring with at most one differing character. The expected output should be 3, as "aba" is the shortest repeating substring with one differing character.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
import sys
input = sys.stdin.readline

def inp():
    return int(input())

def inlt():
    return list(map(int, input().split()))

def insr():
    s = input()
    return list(s[:len(s) - 1])

def invr():
    return map(int, input().split())

def q5(s, n, mid):
    c = 0
    for i in range(mid):
        m = 0
        t = 0
        d = [0] * 26
        for j in range(i, n, mid):
            d[ord(s[j]) - ord('a')] += 1
            m = max(m, d[ord(s[j]) - ord('a')])
            t += 1
        c += (t - m)
    return c <= 1

def main():
    t = inp()
    for _ in range(t):
        n = inp()
        s = input().strip()
        ans = n
        for i in range(1, int(n ** 0.5) + 1):
            if q5(s, n, i):
                ans = min(ans, i)
            if q5(s, n, n // i):
                ans = min(ans, n // i)
        print(ans)

if __name__ == "__main__":
    main()

**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 create a challenging test input that exposes the bug in the provided code, let's focus on edge cases and scenarios where the algorithm might fail to find the correct shortest repeating substring. Here are the steps to consider:

1. **Small Length Strings**: Test strings of length 1 and 2 to ensure the algorithm handles these trivial cases correctly.
2. **Strings with a Single Character**: Test strings where the repeating substring is a single character.
3. **Strings with Repeating Patterns**: Test strings where the repeating pattern is not immediately obvious.
4. **Strings with a Single Mismatch**: Test strings where the repeating pattern has exactly one mismatch.
5. **Strings with Multiple Mismatches**: Test strings where the repeating pattern has more than one mismatch to ensure the algorithm correctly identifies the valid case.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a string with a repeating pattern and a single mismatch.

```plaintext
1
6
abcabc
```

Explanation:
- The string "abcabc" is a repeating pattern of "abc".
- The length of the repeating substring "abc" is 3.
- The string "abcabc" has no mismatches, so the algorithm should correctly identify the length of the shortest repeating substring as 3.

This test case is small and valid, and it should expose any issues in the algorithm's ability to handle repeating patterns correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nearly Shortest Repeating Substring
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $$$s$$$ of length $$$n$$$ consisting of lowercase Latin characters. Find the length of the shortest string $$$k$$$ such that several (possibly one) copies of $$$k$$$ can be concatenated together to form a string with the same length as $$$s$$$ and, at most, one different character.

More formally, find the length of the shortest string $$$k$$$ such that $$$c = \underbrace{k + \cdots + k}_{x\rm\ \text{times}}$$$ for some positive integer $$$x$$$, strings $$$s$$$ and $$$c$$$ has the same length and $$$c_i \neq s_i$$$ for at most one $$$i$$$ (i.e. there exist $$$0$$$ or $$$1$$$ such positions).

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 2\cdot10^5$$$) — the length of string $$$s$$$.

The second line of each test case contains the string $$$s$$$, consisting of lowercase Latin characters.

The sum of $$$n$$$ over all test cases does not exceed $$$2\cdot10^5$$$.

Output: For each test case, print the length of the shortest string $$$k$$$ satisfying the constraints in the statement.

Examples:
input:
5
4
abaa
4
abba
13
slavicgslavic
8
hshahaha
20
stormflamestornflame
output:
1
4
13
2
10

Note: In the first test case, you can select $$$k = \texttt{a}$$$ and $$$k+k+k+k = \texttt{aaaa}$$$, which only differs from $$$s$$$ in the second position.

In the second test case, you cannot select $$$k$$$ of length one or two. We can have $$$k = \texttt{abba}$$$, which is equal to $$$s$$$.

Buggy Code:
from collections import Counter
def solve():
    def ccl(x):
        lst = []
        for i in range(1,int(x ** 0.5) + 1):
            if x % i == 0:
                if i * i == x:
                    lst.append(i)
                else:
                    lst.append(i)
                    lst.append(x // i)
        return lst

    def check(s,idx):
        dir = []
        for i in range(n // idx):
            dir.append(s[i * idx:i * idx + idx])

        if len(set(dir)) > 2:
            return 0
        elif len(set(dir)) == 1:
            return 1
        else:
            dir1 = list(set(dir))
            x1 = dir1[0]
            x2 = dir1[1]
            if idx == 1:
                if s.count(x1) == 1 or s.count(x2) == 1:
                    return 1
                else:
                    return 0
            cnt = 0
            for i in range(idx):
                if x1[i] != x2[i]:
                    cnt += 1
                if cnt > 1:
                    return 0
            else:
                return 1

    n = int(input())
    s = input()
    lst = sorted(ccl(n))
    cnt = len(set(list(s)))
    for i in lst:
        if i >= cnt - 1 and check(s,i):
            print(i)
            break

t = int(input())
for _ in range(t):
    solve()

**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
4
aaaa
5
ababa
12
abcabcabcabc
```