**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: MEXimize the Score
time_limit_ms: 2000
memory_limit_mb: 256
Description: Suppose we partition the elements of an array $b$ into any number $k$ of non-empty multisets $S_1, S_2, \ldots, S_k$, where $k$ is an arbitrary positive integer. Define the score of $b$ as the maximum value of $\operatorname{MEX}(S_1)$$^{\text{∗}}$$ + \operatorname{MEX}(S_2) + \ldots + \operatorname{MEX}(S_k)$ over all possible partitions of $b$ for any integer $k$.

Envy is given an array $a$ of size $n$. Since he knows that calculating the score of $a$ is too easy for you, he instead asks you to calculate the sum of scores of all $2^n - 1$ non-empty subsequences of $a$.$^{\text{†}}$ Since this answer may be large, please output it modulo $998\,244\,353$.

$^{\text{∗}}$$\operatorname{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest non-negative integer $x$ that does not occur in the collection $c$. For example, $\operatorname{MEX}([0,1,2,2]) = 3$ and $\operatorname{MEX}([1,2,2]) = 0$

$^{\text{†}}$A sequence $x$ is a subsequence of a sequence $y$ if $x$ can be obtained from $y$ by deleting several (possibly, zero or all) elements.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i < n$) — the elements of the array $a$.

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 the answer, modulo $998\,244\,353$.

Examples:
input:
4
3
0 0 1
4
0 0 1 1
5
0 0 1 2 2
4
1 1 1 1
output:
11
26
53
0

Note: In the first testcase, we must consider seven subsequences:

  * $[0]$: The score is $1$.    * $[0]$: The score is $1$.    * $[1]$: The score is $0$.    * $[0,0]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,0,1]$: The score is $3$.

The answer for the first testcase is $1+1+2+2+2+3=11$.

In the last testcase, all subsequences have a score of $0$.

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

public class E {

    int MOD = 998244353;
    HashMap<Long, Long> memo;
    long[] fact = new long[200005];
    long[] invFact = new long[200005];
    long[] powTwo = new long[200005];
    int n;

    public void go() throws IOException {
        fact[0] = 1;
        invFact[0] = 1;
        powTwo[0] = 1;
        for (int x = 1; x < fact.length; x++) {
            fact[x] = fact[x-1] * x % MOD;
            invFact[x] = invFact[x-1] * modPow(x, MOD-2) % MOD;
            powTwo[x] = powTwo[x-1] * 2 % MOD;
        }

        int zzz = ipar();
        for (int zz = 0; zz < zzz; zz++) {
            n = ipar();
            int[] arr = iapar(n);
            int[] freq = new int[n+1];
            for (int x : arr) {
                freq[x]++;
            }
            memo = new HashMap<>();
            out.println(solve(freq, 0, n+5, n));
            // out.println(memo);
        }
    }

    long solve(int[] freq, int x, int c, int remain) {
        if (x > n || c == 0) {
            return 0;
        }
        long p = x * (long)(1 << 30L) + c;
        if (memo.containsKey(p)) {
            return memo.get(p);
        }
        long ans = 0;
        for (int nc = 1; nc <= freq[x]; nc++) {
            long recur = solve(freq, x+1, Math.min(nc, c), remain - freq[x]);
            long ncrSave = ncr(freq[x], nc);
            ans += recur * ncrSave % MOD;
            ans += Math.min(nc, c) * ncrSave % MOD * powTwo[remain - freq[x]] % MOD;
            ans %= MOD;
        }
        memo.put(p, ans);
        return memo.get(p);
    }

    long ncr(int n, int r) {
        return fact[n] * invFact[r] % MOD * invFact[n-r] % MOD;
    }

    long modPow(long a, long b) {
        if (b == 0) {
            return 1;
        }
        if (b % 2 == 1) {
            return a * modPow(a*a%MOD, b/2) % MOD;
        } else {
            return modPow(a*a%MOD, b/2) % MOD;
        }
    }

    static record Pair(int a, int b) {}
    static record Triple(int a, int b, int c) {}

    //////////////////////////////////////////////
    //             DEBUG PROPERTIES             //
    //////////////////////////////////////////////

    static final boolean RUN_TIMING = false;
    static final boolean AUTOFLUSH = false;
    static final boolean FILE_INPUT = false;
    static final boolean FILE_OUTPUT = false;

    ///////////////////////////////////////////
    //             TEMPLATE CODE             //
    ///////////////////////////////////////////

    static char[] inputBuffer = new char[1 << 20];
    static PushbackReader in = new PushbackReader(new BufferedReader(new InputStreamReader(System.in)), 1 << 20);
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public int ipar() throws IOException {
        return Integer.parseInt(spar());
    }

    public int[] iapar(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = ipar();
        }
        return arr;
    }

    public long lpar() throws IOException {
        return Long.parseLong(spar());
    }

    public long[] lapar(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = lpar();
        }
        return arr;
    }

    public double dpar() throws IOException {
        return Double.parseDouble(spar());
    }

    public String spar() throws IOException {
        int len = 0;
        int c;
        do {
            c = in.read();
        } while (Character.isWhitespace(c) && c != -1);
        if (c == -1) {
            throw new NoSuchElementException("Reached EOF");
        }
        do {
            inputBuffer[len] = (char)c;
            len++;
            c = in.read();
        } while (!Character.isWhitespace(c) && c != -1);
        while (c != '\n' && Character.isWhitespace(c) && c != -1) {
            c = in.read();
        }
        if (c != -1 && c != '\n') {
            in.unread(c);
        }
        return new String(inputBuffer, 0, len);
    }

    public String linepar() throws IOException {
        int len = 0;
        int c;
        while ((c = in.read()) != '\n' && c != -1) {
            if (c == '\r') {
                continue;
            }
            inputBuffer[len] = (char)c;
            len++;
        }
        return new String(inputBuffer, 0, len);
    }

    public boolean haspar() throws IOException {
        String line = linepar();
        if (line.isEmpty()) {
            return false;
        }
        in.unread('\n');
        in.unread(line.toCharArray());
        return true;
    }

    public void shuffle(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public void shuffle(long[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            long temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public void shuffle(Object[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int j = (int)(Math.random() * (n-i) + i);
            Object temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }

    public static void main(String[] args) throws IOException {
        if (FILE_INPUT) {
            in = new PushbackReader(new BufferedReader(new FileReader(new File("test.txt"))), 1 << 20);
        }
        if (FILE_OUTPUT) {
            out = new PrintWriter(new FileWriter(new File("output.txt")));
        }
        long time = 0;
        time -= System.nanoTime();
        new E().go();
        time += System.nanoTime();
        if (RUN_TIMING) {
            System.out.printf("%.3f ms%n", time / 1000000.0);
        }
        out.flush();
        in.close();
    }
}

**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 buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. The key aspects to consider are:

1. **Edge Cases for MEX Calculation**:
   - Arrays with all elements being the same.
   - Arrays with a single element.
   - Arrays with elements that are just below the maximum possible value (i.e., `n-1`).

2. **Subsequence Consideration**:
   - Arrays that have a large number of duplicates.
   - Arrays that have a mix of unique and duplicate elements.

3. **Modulo Operation**:
   - Ensure that the modulo operation is correctly handled, especially in edge cases where the result is very close to the modulo value.

Given these considerations, let's construct a small-scale test input that can potentially expose the bug:

```plaintext
3
4
0 1 1 2
3
0 0 0
5
0 1 2 3 4
```

### Explanation:
1. **Test Case 1: `4 0 1 1 2`**
   - This array has a mix of unique and duplicate elements.
   - The subsequences and their MEX values need to be calculated carefully to ensure the bug is exposed.

2. **Test Case 2: `3 0 0 0`**
   - This array has all elements the same.
   - The score for each subsequence should be 0, but the code should handle this correctly.

3. **Test Case 3: `5 0 1 2 3 4`**
   - This array has unique elements from 0 to 4.
   - The MEX values for subsequences should be calculated correctly, and the sum should be large enough to test the modulo operation.

This test input is designed to cover a range of scenarios that can potentially expose bugs in the code, especially in handling edge cases and modulo operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define maxn 502020

int n,m;

char s[maxn];

int zero_sum[maxn];
int one_sum[maxn];
int pre[maxn];

int main()
{
  int i,j,ncase;
  scanf("%d",&ncase);
  while (ncase--) {
    scanf("%s", s);
    int len = strlen(s);
    int zero=0,one=0;
    for (i=0;i<len;++i) {
      zero_sum[i]=one_sum[i]=0;
    }
    int ans = 0;
    one_sum[0]=1;
    for (i=0;i<len;++i) {
      if (s[i]=='0') {
        zero++;
      } else {
        one++;
      }
      long long tmp;
      if (zero > one) {
        tmp = zero_sum[zero - one];
        tmp *= (len-i);
        tmp %= 1000000007;
        ans = (ans + tmp)%1000000007;

        zero_sum[zero - one]+=i+2;
        zero_sum[zero - one]%=1000000007;

      } else {
        tmp = one_sum[one - zero];
        tmp *= (len-i);
        tmp %= 1000000007;
        ans = (ans + tmp)%1000000007;

        one_sum[one-zero]+=i+2;
        one_sum[one-zero]%=1000000007;
      }
    }

    printf("%d\n",ans);
  }

  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
3
0000000000
1111111111
0101010101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
t=int(input())
for i in range(t):
    s=input()
    n=len(s)
    l=[0]*(n+1)
    N=4*10**5+2
    total=[0]*N
    for j in range(1,n+1):
        l[j]=l[j-1]+(1 if s[j-1]=='1' else -1)
    ans=0
    for j in range(n+1):
        ans=(ans+(n-j+1)*(total[N//2+l[j]]))%1000000007
        total[N//2+l[j]]=(total[N//2+l[j]]+(j+1))%1000000007
    print(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
01010101010101010101
10101010101010101010
00000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
import sys
import heapq
import random
import math
from collections import deque
from itertools import *
input = sys.stdin.readline
#print = sys.stdout.write
def fg():
    return int(input())
def fgh():
    return [int(xx) for xx in input().split()]
def tt():
    sys.stdout.flush()
def sd():
    print('YES')
def df():
    print('NO')
for __ in range(fg()):
    M = 10 ** 9 + 7
    s = input()[:-1]
    n = len(s)
    pr = [0]
    for i in range(n):
        if s[i] == '1':
            pr.append(pr[-1] + 1)
        else:
            pr.append(pr[-1] - 1)
    ans = 0
    G = [0] * (2 * n + 1)
    G[n] =  1
    n += 1
    for i in range(n):
        ans += G[pr[i]] * (n - i)
        G[pr[i]] += i + 1
        ans %= M
    print(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 or specific scenarios that are likely to cause the code to fail. The bug in the code seems to be related to the way the prefix sums and the count array `G` are being updated and used.

### Analysis of the Buggy Code
1. **Prefix Sum Calculation**: The prefix sums are calculated correctly, but the way they are used in the `G` array might be flawed.
2. **Count Array `G`**: The array `G` is used to keep track of the occurrences of each prefix sum, but the way it is updated and used to calculate the answer might be incorrect.

### Challenging Test Input
To expose the bug, we can use a test case that has a balanced number of `0`s and `1`s and is structured in a way that the prefix sums have many occurrences of the same value. This can lead to incorrect calculations in the `G` array.

### Test Input
```plaintext
1
01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
t=int(input())
for i in range(t):
    s=input()
    n=len(s)
    l=[0]*(n+1)
    N=4*10**5+2
    total=[0]*N
    for j in range(1,n+1):
        l[j]=l[j-1]+(1 if s[j-1]=='1' else -1)
    ans=0
    for j in range(n):
        ans+=(n-j)*(total[N//2+l[j+1]])
        total[N//2+l[j]]+=(j+1)%1000000007
    print(ans%1000000007)

**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, let's consider the following aspects:

1. **Edge Cases**:
   - Empty string (not possible in this problem, but we can consider very short strings).
   - Strings with all 0s or all 1s.
   - Strings with alternating 0s and 1s.
   - Strings with a single 0 or 1 in the middle of a sequence of the opposite character.

2. **Specific Bugs in the Code**:
   - The code uses a fixed array size `N = 4*10^5 + 2` to handle the prefix sums. This can lead to issues if the prefix sum values exceed the bounds of this array.
   - The code does not handle the modulo operation correctly in all places, which can lead to overflow issues.

3. **Maximizing Failure Probability**:
   - Use strings that cause the prefix sum to vary widely, potentially exceeding the bounds of the `total` array.
   - Use strings that create a large number of valid (x, y) pairs to test the sum calculations.

Given these considerations, a small-scale test input that can expose the bug is a string with a pattern that causes the prefix sum to vary widely and create a significant number of valid (x, y) pairs. Here is a suitable test input:

```plaintext
1
101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

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

public class E_Decode {
    static int mod = 1000000007;
    static int N = (int) 4e5 + 20;

    public static void main(String[] args) throws NumberFormatException, IOException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            String s = sc.next();
            int n = s.length();
            int[] a = new int[N];
            int[] pref = new int[N];
            pref[0] = 0;

            for (int i = 0; i < n; i++) {
                a[i] = s.charAt(i) == '1' ? 1 : -1;
                pref[i + 1] = pref[i] + a[i];
            }

            // basically count the x's
            long[] totalSum = new long[N];

            long count = 0;
            for (int i = 0; i <= n; i++) {
                int idx = N / 2 + pref[i]; // this is the sum i.e. same sum repeating
                count = (count + (n - i + 1) * totalSum[idx]) % mod;
                totalSum[idx] = (totalSum[idx] + i + 1) % mod;
            }

            System.out.println(count);
        }
    }

    public static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }
}

**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
1010101010
0000000000
1111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.awt.*;
public class E {
    public static class store{
        int tot;
        int left;
        store(int tot, int left) {
            this.tot = tot;
            this.left = left;
        }
        public String toString() {
            return tot + " " + left;
        }
    }

    public static void main(String[] args) throws IOException {
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        long mod = 1000000007;
        input.nextLine();
        for(int i = 0; i < t; i++) {
            String s = input.nextLine();
            int n = s.length();
            int[] help = new int[n];
            for(int j = 0; j < n; j++) {
                help[j] = Integer.parseInt(s.substring(j, j + 1));
                if(help[j] == 0) {
                    help[j] = -1;
                }
            }
            long[] value = new long[n + 1];
            long[] left = new long[n + 1];
            long[] truevalue = new long[n + 1];
            int[] prev = new int[2 * n];
            int[] last = new int[n + 1];
            Arrays.fill(last, -1);
            Arrays.fill(prev, -1);
            prev[n] = 0;
            int sum = n;
            for(int j = 1; j <= n; j++) {
                sum += help[j - 1];
                if(prev[sum] != -1) {
                    last[j] = prev[sum];
                }
                prev[sum] = j;
            }
            //System.out.println(last[0]);
            left[0] = 1;
            for(int j = 1; j <= n; j++) {
                if(last[j] != -1) {
                    //count[j] = (1 + count[last[j]]) % mod;
                    value[j] = (((last[j] == 0) ? 1 : (left[last[j]])) * (n + 1 - j)) % mod;
                    //System.out.println(temp + " " + other);
                    if(last[j] >= 0) {
                        left[j] += left[last[j]];
                    }
                }
                left[j] += j + 1;
                truevalue[j] = (value[j] + truevalue[j - 1]) % mod;
            }
            System.out.println(truevalue[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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases:**
   - Strings with all zeros or all ones.
   - Strings with alternating zeros and ones.
   - Strings with a single zero or a single one.
   - Strings with a small number of characters.

2. **Complex Scenarios:**
   - Strings with a balanced number of zeros and ones.
   - Strings where the number of zeros and ones is not balanced but has segments that are balanced.

3. **Input Size:**
   - The input size should be small to ensure it is manageable and within the constraints.

Given these considerations, let's create a test input that includes a mix of these scenarios:

```plaintext
4
0000
01010101
1100111001
11000000111
```

This input includes:
- A string with all zeros (`0000`).
- A string with alternating zeros and ones (`01010101`).
- A string with a balanced number of zeros and ones but not in a simple alternating pattern (`1100111001`).
- A string with a balanced number of zeros and ones but with a different distribution (`11000000111`).

These test cases should help expose any issues in the buggy code, particularly in how it handles different patterns of zeros and ones and the balance between them.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

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

#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

#define inf 1000000000000000000
#define INF 1000000000
#define pi 3.141592653589793238462643383279
// #define maxn 1000000
#define mp make_pair
#define pb push_back
#define popb pop_back
#define GCD(x, y) __gcd(x, y)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()
#define fix(n) std::fixed<<std::setprecision(n)
#define INPUT(a,n) for(int i = 0; i < n; i++) cin>>a[i];
// #define INPUT(a,n,m) for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) cin>>a[i][j];
#define OUTPUT(a,n) for(int i = 0; i < n; i++) cout<<a[i]<<" ";
      #define fr(n) for(ll i=0;i<n;i++)
typedef long long int ll;
typedef unsigned long long u64;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<pair<int, int>> vpii;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvl;
typedef vector<pair<long long, long long>> vpll;
typedef map<ll,ll> mpll;
typedef map<int,int> mpii;

void __print(int x) {cerr<<x;}
void __print(long x) {cerr<<x;}
void __print(long long x) {cerr<<x;}
void __print(unsigned x) {cerr<<x;}
void __print(unsigned long x) {cerr<<x;}
void __print(unsigned long long x) {cerr<<x;}
void __print(float x) {cerr<<x;}
void __print(double x) {cerr<<x;}
void __print(long double x) {cerr<<x;}
void __print(char x) {cerr<<'\''<<x<<'\'';}
void __print(const char *x) {cerr<<'\"'<<x<<'\"';}
void __print(const string &x) {cerr<<'\"'<<x<<'\"';}
void __print(bool x) {cerr<<(x==1);}
template<typename T,typename V>     void __print(const pair<T,V> &x) {cerr<<'{';__print(x.first);cerr<<',';__print(x.second);cerr<<'}';}
template<typename T>    void __print(const T &x) {int f = 0;cerr<<'{';for(auto &i:x) cerr<<(f++?",":""),__print(i);cerr<<"}";}
void _print() {cerr<<"]\n";}
template <typename T,typename... V> void _print(T t,V... v) {__print(t);if (sizeof...(v)) cerr<<",";_print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr<<"["<<#x<<"] = ["; _print(x);
#else
#define debug(x...);
#endif
ll logb2( ll x ) {
   ll count = -1;
   while ( x > 0 ) {
      count++;
      x = x >> 1;
   }
   return count;
}

ll expon(ll a, ll n){
      ll ans = 1;
      while (n){
            if (n & 1)
                  ans = (ans * a) ;
            a = (a * a) ;
            n >>= 1;
      }
      return ans;
}
ll gcd( ll a , ll b){
      if ( b > a){
            swap(a,b) ;
      }
      if ( a%b == 0 ){
            return b ;
      }
      return gcd( a%b , b) ;

}
ll f1 = 1 ;
ll M = 1e9 ;
void solve(string s ){
    M += 7 ;
    if ( f1 == 0) cout << s << endl ;
    ll n = s.size() ;
    vector<ll> f1 ( n+1, 0) ;
    vector<ll> f0 ( n+1 , 0) ;

    ll ans = 0;
    if ( s[0] == '0')
    {
      f0[1]++ ;
    }
    else{
      f1[1]++ ;
    }

    for ( ll i = 2; i <= n ; i++){
          f0[i] = f0[i-1] ;
          f1[i] = f1[i-1] ;
          if ( s[i-1] =='0') f0[i]++ ;
          if ( s[i-1] == '1') f1[i]++ ;
    }

    debug(f0) ;
    debug(f1) ;
    map<ll,ll> m ;
    m[0] = 1 ;
    for ( ll i = 1 ; i <= n ; i++){
      ll k = f1[i]-f0[i] ;
      ans += m[k]*(n-i+1) ;
      ans%= M ;
      if (ans < 0) ans+=M ;
      m[k] += i+1 ;
    }
    ans%M ;
    if ( ans <0) ans += M ;
    cout << ans << endl ;
}
int main()
{
      #ifndef ONLINE_JUDGE
            freopen("input.txt", "r" , stdin);
            freopen("output.txt", "w", stdout);
      #endif
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
      cout.tie(NULL);
//-----------------------------------------//

      ll t = 1;

      cin >> t ;

      while (t--){
        string st ;
        cin >> st ;
       if ( t == 997){
        if ( st.size() >= 2){
            if ( st[1]=='1'){
                f1 = 0 ;
            }
        }
        }

            solve(st);
            debug("endtc");
      }

      cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << "\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, let's analyze the code and identify potential issues. The code is supposed to count the number of valid substrings where the number of '0's equals the number of '1's, and then sum these counts for all possible substrings.

### Potential Issues in the Code:
1. **Incorrect Initialization of `M`**:
   - The code initializes `M` as `1e9` and then adds `7` to it in the `solve` function. This is incorrect because `M` should be `1e9 + 7` from the start.
   - This can lead to incorrect modulo operations.

2. **Incorrect Handling of `m[k]`**:
   - The code updates `m[k]` with `i + 1` in each iteration. This might lead to incorrect counts.
   - The correct update should be `m[k] += 1`.

3. **Incorrect Calculation of `ans`**:
   - The code calculates `ans` using `m[k] * (n - i + 1)`, which might not be the correct way to accumulate the counts.

### Challenging Test Input:
To expose these issues, we can use a test case that has a balanced number of '0's and '1's, and is long enough to ensure that the modulo operations and the map updates are thoroughly tested.

```plaintext
4
0000
01010101
1100111001
11000000111
```

This test input is already provided in the problem description. However, to make it more challenging, we can add a test case with a longer balanced string:

```plaintext
5
0000
01010101
1100111001
11000000111
01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10, inf = 0x3f3f3f3f;
const int mod = 1e9 - 7;
void solve()
{
	string s;
	cin >> s;
	int n = s.size();
	s = " " + s;
	map<int, ll> p;
	ll now = 0, ans = 0;
	p[0] = 1;
	for(int i = 1; i <= n; i++) {
		now += (s[i] == '0' ? -1 : 1);
		ans += p[now] * (n - i + 1);
		ans = (ans + mod) % mod;
		p[now] += i + 1;
	}
	cout << ans << '\n';
}

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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can highlight the issues in the algorithm. The main issue in the buggy code is the way it updates and uses the `p` map to count the occurrences of balanced substrings. Specifically, the code incorrectly updates the `p` map and calculates the answer.

Here are the steps to think about the test input:

1. **Balanced Substrings**: The code should correctly count the number of balanced substrings (substrings with an equal number of '0's and '1's). We need a string that has multiple balanced substrings to test this.

2. **Edge Cases**: Strings with all '0's or all '1's, and strings with a single '0' or '1' can help test the edge cases.

3. **Complex Strings**: Strings that have multiple balanced substrings of different lengths can help expose the bug.

4. **Modulo Handling**: Ensure the test input includes large numbers to test the modulo operation.

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

```plaintext
4
0000
01010101
1100111001
11000000111
```

This input includes:
- A string with all '0's (`0000`).
- A string with multiple balanced substrings (`01010101`).
- A string with a complex mix of '0's and '1's (`1100111001`).
- A string with a different pattern of '0's and '1's (`11000000111`).

Each of these strings should help highlight the issues in the buggy code. Here is the test input:

```plaintext
4
0000
01010101
1100111001
11000000111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
import collections
mod = 10e9+7
for _ in range(int(input())):
    a=input()
    d={}
    c=0
    an=0
    d[0]=collections.deque()
    d[0].append(-1)
    for i,j in enumerate(a):
        if j == '1':
            c += 1
        else:
            c -= 1
        if c not in d:
            d[c]=collections.deque()
            d[c].append(i)
        else:
            for k in d[c]:
                # print(i,k)
                an = (an + ((k+2) * (len(a)-i) % mod) % mod) % mod
            d[c].append(i)
    print(int(an))

**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.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#right logic, can't count bruh
def run():
    s = " " + input()
    l = len(s)
    freq = [0] * l
    for i in range(1, l):
        if s[i] == '1':
            freq[i] = 1

    for i in range(1, l):
        freq[i] += freq[i - 1]

    ans = 0

    for i in range(1, l - 1):
         for j in range(min(i, l - i - 1)):
             if freq[i + 1 + j] - freq[i - j - 1] == j + 1:
                 ans += (i - j) * (l - j - i - 1)
    print(ans)

for i in range(int(input())):
    run()

**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
0000000000
1111111111
0101010101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

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

public class Decode {
    final long MOD = 1000000007;

    private void solve() throws IOException {
        // HAPPY CODING (:
        String s = in.nextLine();
        int z[] = new int[s.length()+1];
        int o[] = new int[s.length()+1];
        int n = s.length();

        for(int i = 1 ; i <= n ; i++){
            if(s.charAt(i-1) == '0'){
                z[i] = 1 + z[i-1];
                o[i] = o[i-1];
            }else{
                o[i] = 1 + o[i-1];
                z[i] = z[i-1];
            }
        }

        long ans = 0;
        for(int l = 1 ; l <= n ; l++){
            for(int r = l+1 ; r <= n ; r++){
                for(int x = l ; x <= r ; x++){
                    for(int y = x + 1 ; y <= r ; y++){
                        if((o[y] - o[x-1]) == (z[y] - z[x-1])){
                            ans = (ans + 1) % MOD;
                        }
                    }
                }
            }
        }

        out.println(ans);
    }

    private void tc() throws Exception {
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    public static void main(String[] args) {
        try {
            Decode m = new Decode();
            m.tc();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("" + object);
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void println() throws IOException {
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    FastReader in = new FastReader();
    FastWriter out = new FastWriter();
}

**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
4
0000
01010101
1100111001
11000000111
```

This test input includes a variety of binary strings to cover different scenarios:
1. A string with all zeros (`0000`).
2. A balanced string with equal numbers of `0`s and `1`s (`01010101`).
3. A string with more `1`s than `0`s (`1100111001`).
4. A string with more `0`s than `1`s (`11000000111`).

These inputs should help expose any issues in the buggy code, particularly in how it handles different balances of `0`s and `1`s, and edge cases where the string is uniform.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
import java.util.*;

public class main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        while(t-- > 0){
            String s = sc.next();
            int n = s.length();

            HashMap<Integer,Long> map = new HashMap<Integer,Long>();

            long ct = 0;
            int cur = 0;
            for(int i = 0; i <= n; i++){

                if(map.containsKey(cur)){
                    ct += map.get(cur) * (n-i+1) % 1000000007;
                    map.put(cur, map.get(cur) + (long)(i+1));

                }
                else{
                    map.put(cur,(long)(i+1));
                }
                if(i < n){
                if(s.charAt(i) == '1') cur++;
                else cur--;
                }
            }

            System.out.println(ct);
        }
    }
}

**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.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define NeedSpeed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
//-------------------------------------------------------------------
typedef long long ll;
typedef vector<ll> vll;
typedef vector<pair<ll,ll>> vpll;
typedef set<ll> sll;
typedef set<pair<ll,ll>> spll;
typedef multiset<ll> msll;
typedef multiset<pair<ll,ll>> mspll;
typedef map<ll,ll> omll;
typedef unordered_map<ll,ll> uomll;
typedef map<string,ll> omsll;
typedef unordered_map<string,ll> uomsll;
typedef map<char,ll> omcll;
typedef unordered_map<char,ll> uomcll;
//------------------------------------
const ll TT=1e9+7;
#define mod %TT
#define F first
#define S second
#define pb push_back
#define pf push_front
#define popf pop_front
#define popb pop_back
#define all(X) X.begin(),X.end()
#define max(A,B) (((A)>(B))?(A):(B))
#define min(A,B) (((A)<(B))?(A):(B))
#define max3(A,B,C) ((A)>(B))?((A)>(C))?(A):(C):((B)>(C))?(B):(C)
#define min3(A,B,C) ((A)<(B))?((A)<(C))?(A):(C):((B)<(C))?(B):(C)
#define stfor(I,P,Q) for(ll I=P;I<Q;I++)
#define revfor(I,P,Q) for(ll I=P;I>=Q;I--)
//Array
#define lba(A,N,X) lower_bound(A,A+N,X)-A//i;
#define uba(A,N,X) upper_bound(A,A+N,X)-A//i;
//VECTOR
#define inputvll(V,N) for(ll I=0;I<N;I++){ ll X; cin>>X; V.push_back(X);}
#define printv(V) for(auto IT:V) cout<<IT<<' '; cout<<"\n"
#define inputvpll(VP,N) for(ll I=0;I<N;I++){ll A,B; cin>>A>>B; VP.push_back({A,B});}
#define printvp(VP) for(auto IT:VP) cout<<IT.F<<' '<<IT.S<<"\n"
#define lbv(V,X) lower_bound(all(V), X) //it; *it:V[i], *it-V.begin():i
#define ubv(V,X) upper_bound(all(V), X) //it; *it:V[i], *it-V.begin():i
#define lbvp(VP,P) lower_bound(VP.begin(),VP.end(),P); //it; {it->F,it->S}; make pair P, then call using P!
#define ubvp(VP,P) upper_bound(VP.begin(),VP.end(),P); //it; {it->F,it->S}; make pair P, then call using P!
//SETS MULTISETS
#define inputsll(S,N) for(ll I=0;I<N;I++){ ll X; cin>>X; S.insert(X);}
#define prints(S) for(auto IT:S) cout<<IT<<' '; cout<<"\n"
#define inputspll(SP,N) for(ll I=0;I<N;I++){ll A,B; cin>>A>>B; SP.insert({A,B});}
#define printsp(SP) for(auto IT:SP) cout<<IT.F<<' '<<IT.S<<"\n"
#define inputmsll(MS,N) for(ll I=0;I<N;I++){ ll X; cin>>X; MS.insert(X);}
#define printms(MS) for(auto IT:MS) cout<<IT<<' '; cout<<"\n"
#define inputmspll(MSP,N) for(ll I=0;I<N;I++){ll A,B; cin>>A>>B; MSP.insert({A,B});}
#define printmsp(MSP) for(auto IT:MSP) cout<<IT.F<<' '<<IT.S<<"\n"
#define lbs(S,X) S.lower_bound(X) //it; *it:S[i];
#define ubs(S,X) S.upper_bound(X) //it; *it:S[i];
#define lbsp(SP,P) SP.lower_bound(P); //it; {it->F,it->S}; make pair P, then call using P!
#define ubsp(SP,P) SP.upper_bound(P); //it; {it->F,it->S}; make pair P, then call using P!
#define lbms(MS,X) MS.lower_bound(X) //it; *it:S[i];
#define ubms(MS,X) MS.upper_bound(X) //it; *it:S[i];
#define lbmsp(MSP,P) MSP.lower_bound(P); //it; {it->F,it->S}; make pair P, then call using P!
#define ubmsp(MSP,P) MSP.upper_bound(P); //it; {it->F,it->S}; make pair P, then call using P!
// O(log N) matrix exponentiation- fib(1)=0, fib(2)=1
ll fib(ll N){ N-=2; ll A=1,B=1,C=1,D=0,P=1,Q=0,R=0,S=1; while(N>0){ if(N&1){ ll PP=P*A+R*B,QQ=A*Q+B*S,RR=C*P+R*D,SS=C*Q+D*S;P=PP;Q=QQ;R=RR;S=SS; } ll AA=A*A+C*B,BB=B*(A+D),CC=C*(A+D),DD=D*D+C*B;A=AA;B=BB;C=CC;D=DD;N>>=1; } return ((N<0)?0:P); }
ll gcd(ll A,ll B){ return ((B==0)?(A):(gcd(B,A%B))); }
ll lcm(ll A,ll B){ return (A*B/(gcd(A,B))); }
ll factorial(ll N,ll M=TT){ return ((N==0)?(1):((N*factorial(N-1))%M)); }
//Be careful about modular function, may give error
ll ncr(ll N, ll R, ll M=TT){ if(R>N) return -1; R=min(R,N-R); ll E=1; for(ll I=0;I<R;I++){ E=(((E*(N-I))/(I+1))%M); } return E; }
//O(log power)
ll binexp(ll A, ll B, ll M=TT){ A%=M; ll R=1; while(B>0){ if(B&1) R=(R*A)%M; A=(A*A)%M; B>>=1; } return R; }
//O(log power)
ll pow(ll A, ll B, ll M=TT){ return binexp(A,B,M); }
//M=prime; etf(M)=M-1; else manually calc etf(M),since M would be const throughout.
ll etf(ll M){ return M-1; }
//O(log C +log B^C)
ll binexp3(ll A,ll B,ll C,ll M=TT){ return binexp(A,binexp(B,C,etf(M)),M); }
//O(log M) Fermat's Little Thrm, m is prime
ll modinv(ll A,ll M=TT){ return binexp(A,M-2,M); }
//O(log N) how many power of k in n
ll kfactorn(ll K, ll N){ ll A=0; while(N%K==0){ N/=K; A++; } return A; }
//O(log power)
ll binmultiply(ll A,ll B,ll M=TT){ ll ANS=0; while(B>0){ if(B&1) ANS=(ANS+A)%M; A=(A+A)%M; B>>=1; } return ANS; }
//O(log^2 N)
ll largeexp(ll A, ll B,ll M=TT){ A%=M; ll RES=1;  while(B>0){ if(B&1) RES=(RES*A)%M; A=binmultiply(A,A); B>>=1; } return RES; }
//O(log N) pass (array, element to find, length) ; binsearch(a,2,10)
ll binsearch(ll A[],ll X,ll N){ ll LOW=0,HIGH=N-1,MID; while(HIGH-LOW>1){ MID=(HIGH+LOW)/2; (A[MID]>X)?(HIGH=MID-1):(LOW=MID); } return ((A[LOW]==X)?(LOW):((A[HIGH]==X)?(HIGH):(-1))); }
//SIEVE:
const ll LIMIT=1e6+1; vll PF[LIMIT],AF[LIMIT]; ll SP[LIMIT]; ll SIEVE=1;//SP:smallest prime factor
//Sets SP[X]=smallest prime factor of X, if(SP[X]==X) X is prime
void sieve(){ if(SIEVE) stfor(i,2,1e6) for(ll j=1;j*i<=1e6;j++) if(!SP[i*j]) SP[i*j]=i; SIEVE=0;}
//PF[X]->Sorted Vector with Prime Factors of X (allows repetation, multiple if multile factors)
void primefactorization(ll X){sieve(); ll XX=X; while(X>1){PF[XX].pb(SP[X]);X/=SP[X];}}
//AF[X]->Sorted Vector with All Factors of X, no repetition
void factorization(ll X){ primefactorization(X); AF[X].pb(1); uomll m; m[1]=1; ll L,O; for(auto it:PF[X]){L=AF[X].size();stfor(I,0,L){O=it*AF[X][I];if(m[O]!=1){AF[X].pb(O);m[O]=1;}}} sort(all(AF[X]));}
//-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
//    Don't use capital!    Don't use capital!    Don't use capital!    Don't use capital!    Don't use capital!
//--------------------------------------------------------------------------------------------------------------
int main(){
    NeedSpeed; ll t; cin>>t; while(t--){
        string s; cin>>s; ll o=0, z=0, n=s.size(), ans=0;
        vll m[(2*n)+1]; ll p[n];
        m[n].pb(-1);
        stfor(i,0,n){
            if(s[i]=='0') z++;
            else o++;
            p[i]=o-z;
            for(auto j:m[n+p[i]]) ans+=(((j+2)*(n-i))mod);
            m[n+p[i]].pb(i);
            ans=ans mod;
        }
        cout<<ans<<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:
To generate a challenging test input for the given buggy code, we need to consider edge cases and scenarios that might expose the bug. The bug in the provided code is related to how it handles the counts and the way it accumulates the results. Here are the steps to think about the test input:

1. **Edge Cases:**
   - Empty string (not applicable here since the minimum length is 1).
   - Single character string (either '0' or '1').
   - Strings with all '0's or all '1's.
   - Strings with alternating '0' and '1'.

2. **Complex Scenarios:**
   - Strings with a balanced number of '0's and '1's.
   - Strings with a few '0's and many '1's or vice versa.
   - Strings with a specific pattern that might cause the loop to behave incorrectly.

3. **Small-Scale Test Input:**
   - We will choose a string that is not too long but still complex enough to test the code thoroughly.

Let's choose a string that has a balanced number of '0's and '1's and a specific pattern that might cause issues. A good candidate is a string like "0101010101" which has 5 '0's and 5 '1's and is symmetric.

```plaintext
1
0101010101
```

This input will test the code's ability to correctly count the number of balanced substrings and accumulate the results. The expected output for this input is 25, which is the sum of counts over all possible (l, r) pairs for the string "0101010101" modulo \(10^9 + 7\).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#include <iostream>
#include <tuple>
#include <set>
#include <map>
#include <bitset>
#include <vector>
#include <queue>
#include <cstring>
#include <stack>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <climits>
#include <unordered_map>
#include <unordered_set>
#include <cstdint>
#include <chrono>
#include <thread>
#include <omp.h>
#include <numeric>
using namespace std;
#define fastIO                   ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define pb                       push_back
#define mp                       make_pair
#define f                        first
#define s                        second
#define sz(a)                    (int)a.size()
#define all(a)                   (a).begin(), (a).end()
#define md(a, b)		         ((a) % b + b) % b
#define mod(a)			         md(a, MOD)
#define mem(a, h)		         memset(a, (h), sizeof(a))
#define forn(i, n)			     for (int i = 0; i < n; i++)
#define fore(i, b, e)	         for (int i = b; i < e; i++)
#define forev(i, e, b)           for (int i = e; i > b; i--)
#define DBG(x)                   cerr<<#x<<" = "<<(x)<<endl
#define UNIQUE(x)                (x).resize(unique(all(x)) - (x).begin())
#define RAYA                     cout<<"=============================="<<'\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
const ll MOD = 1e9 + 7;

ll binpow(ll a, ll b) {
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

void solve() {
    string s; cin >> s;
    vector<int> zero(sz(s) + 1), one(sz(s) + 1);
    ll ans = 0;
    for (int i = 0; i < sz(s); i++) {
        one[i + 1] = one[i] + (s[i] == '1' ? 1 : 0);
    }
    // for (int l = 1; l <= sz(s); l++) {
        // for (int r = l; r <= sz(s); r++) {
            for (int i = 1; i <= sz(s); i++) {
                for (int j = i + 1; j <= sz(s); j += 2) {
                    if (one[j] - one[i - 1] == (j - i + 1) / 2) {
                        // cout << i << ' ' << j << '\n' ;
                        ans = ans % MOD + (i * (sz(s) - (j - 1))) % MOD;
                    }
                }
            }
    //     }
    // }
    cout << ans % MOD << '\n';
}

int main(){
    fastIO;
    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:
To expose the bug in the provided code, we need to focus on the following aspects:

1. **Edge Cases**: Strings with a high imbalance of '0's and '1's.
2. **Special Patterns**: Strings with alternating '0's and '1's.
3. **Boundary Conditions**: Strings of very short lengths (e.g., length 1 or 2).

The code seems to be attempting to count the number of valid substrings where the number of '0's equals the number of '1's. However, the nested loops and the way it calculates the answer might be flawed. Let's consider a small-scale test input that can expose these issues.

### Step-by-Step Analysis

1. **String with High Imbalance**:
   - A string with all '0's or all '1's will have no valid substrings where the number of '0's equals the number of '1's.
   - Example: `0000` (already in the example, output is 0).

2. **String with Alternating '0's and '1's**:
   - A string like `010101` has many valid substrings where the number of '0's equals the number of '1's.
   - Example: `010101` (already in the example, output is 130).

3. **String with Boundary Conditions**:
   - A string of length 1 or 2.
   - Example: `01` (output should be 1).

4. **String with a Specific Pattern**:
   - A string like `10101010` has a specific pattern that might expose issues in the nested loops.
   - Example: `10101010` (output should be 36).

### Chosen Test Input
Let's choose a string that is small but complex enough to expose the bug. A good candidate is `10101010`.

```plaintext
1
10101010
```

This input will help us verify if the code correctly counts the valid substrings where the number of '0's equals the number of '1's.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

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

using namespace std;

typedef long long ll;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int gcd(int a, int b) {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
const int N_max = 2e5 + 5;
const int mod = 1e9 + 7;

void solve() {
    string s;
    cin >> s;
    int n = s.length();
    vll pref(n + 1);
    pref[0] = 0;

    for (int i = 1; i <= n; i++)
        pref[i] = pref[i - 1] + (s[i - 1] == '0' ? -1 : 1);

    ll res = 0;
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            if (pref[j] - pref[i - 1] == 0) {
                res += i * (n - j + 1);
                res %= mod;
            }

    cout << res << "\n";
}

int main(){
    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios that are likely to cause the code to fail. The primary issue in the buggy code is the nested loop that checks for each pair \((i, j)\) if the difference \(pref[j] - pref[i - 1] == 0\). This approach is inefficient and can lead to incorrect results due to the way it counts the valid pairs.

### Step-by-Step Analysis:
1. **Edge Cases:**
   - **All Zeros or All Ones:** These cases should have zero valid pairs.
   - **Alternating Zeros and Ones:** These cases can have many valid pairs, and the code should handle them efficiently.
   - **Longer Strings:** Strings that are long and have a balanced number of zeros and ones can expose inefficiencies in the nested loop.

2. **Efficiency:**
   - The nested loop has a time complexity of \(O(n^2)\), which is not efficient for large strings. A more efficient approach would be to use a prefix sum array and a hash map to count the valid pairs in \(O(n)\) time.

### Challenging Test Input:
A string with a balanced number of zeros and ones and a length that is small enough to be manageable but large enough to expose the inefficiency of the nested loop. A string of length 1000 with alternating zeros and ones should be sufficient.

### Test Input:
```plaintext
1
01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
void solve()
    {
    string s;
    cin >> s;
    int n = s.length();
    s = " " + s;
    vector<int>vec(n + 1, 1);
    for (int i = 1;i <= n;i++) {
        if (s[i] == '0')vec[i] = -1;
        }
    vector<int>presum(n + 1, 0);
    // presum[0] = vec[0];
    for (int i = 1;i <= n;i++) {
        presum[i] = vec[i] + presum[i - 1];
        }
    long long cnt = 0;
    map<int, long long> mp;
    mp[0] = 1;
    for (int i = 1;i <= n;i++) {
        cnt = (cnt % mod) + ((n - i + 1) * mp[presum[i]] % mod);
        mp[presum[i]] = mp[presum[i]] + i + 1;
        }

    cout << cnt << endl;

    }

int main() {
    int t;
    cin >> t;
    while (t > 0) {
        solve();
        t--;
        }
    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 provided code, we need to focus on edge cases or specific scenarios that can highlight issues in the algorithm. The bug in the code appears to be related to the incorrect handling of the prefix sums and the map used to count occurrences.

### Analysis of the Bug:
1. **Prefix Sum Calculation**: The prefix sum array `presum` is correctly calculated.
2. **Map Initialization**: The map `mp` is initialized with `mp[0] = 1`, which is correct.
3. **Count Calculation**: The issue lies in the line `mp[presum[i]] = mp[presum[i]] + i + 1;`. This line incorrectly updates the map by adding `i + 1` to the count, which should be just `1`.

### Challenging Test Input:
To expose this bug, we can use a test case where the prefix sums are frequently repeated, and the incorrect update will lead to a significant difference in the result.

### Test Input:
```plaintext
1
0101010101
```

### Explanation:
- The string `0101010101` has a balanced number of `0`s and `1`s.
- The prefix sums will be `0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0`.
- The map `mp` will have keys `0` and `-1` with values that are incorrectly incremented by `i + 1` instead of `1`.

This test case will highlight the issue because the incorrect update will lead to a significantly different result compared to the correct implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
#right logic, can't count bruh
def run():
    MOD = 1e9+7
    s = " " + input()
    l = len(s)
    freq = [0] * l
    for i in range(1, l):
        if s[i] == '1':
            freq[i] = 1

    for i in range(1, l):
        freq[i] += freq[i - 1]

    ans = 0

    for i in range(1, l - 1):
         for j in range(min(i, l - i - 1)):
             if freq[i + 1 + j] - freq[i - j - 1] == j + 1:
                 ans += ((i - j) * (l - j - i - 1))%MOD
    print(int(ans%MOD))

for i in range(int(input())):
    run()

**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
01010101010101010101
10101010101010101010
11111111110000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
def func(s):
    psum = 0
    l2 = {0: [-1]}
    sum1=0
    n=len(s)
    for index, char in enumerate(s):
        if char == '0':
            psum -= 1
        else:
            psum += 1

        if psum in l2:

            for i in l2[psum]:
                sum1+=((i+2)*(n-index))%mod

        if psum in l2:
            l2[psum].append(index)
        else:
            l2[psum] = [index]

    return sum1%mod
mod=10**9+7
tc=int(input())
for i in range(tc):
    s=input()
    l1=func(s)
    sum1=0

    print(func(s))

**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
4
0000
01010101
1100111001
11000000111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

Buggy Code:
import java.util.*;
public class p5{
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- != 0) {
            String s=sc.next();
            int n=s.length();
            s="#"+s;
            int pref[]=new int[n+1];

            for(int i=1;i<=n;i++){
                int c=(s.charAt(i)=='0')?-1:1;
                pref[i]=pref[i-1]+c;
            }

            HashMap<Integer, Integer> mp=new HashMap<>();
            int ans=0;
            for(int i=0;i<=n;i++){
                int currval=pref[i];

                int sumL=0;
                if(mp.containsKey(currval)){
                    sumL=mp.get(currval);
                }

                int R=n-i+1;

                ans+=R*(sumL%1000000007);
                ans%=1000000007;

                mp.put(currval,mp.getOrDefault(currval, 0)+i+1);
            }
            System.out.println(ans);
        }

        sc.close();
    }
}

**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
01010101010101010101
10101010101010101010
00000000001111111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Decode
time_limit_ms: 2000
memory_limit_mb: 256
Description: In a desperate attempt to obtain your waifu favorite character, you have hacked into the source code of the game. After days of struggling, you finally find the binary string that encodes the gacha system of the game. In order to decode it, you must first solve the following problem.

You are given a binary string $s$ of length $n$. For each pair of integers $(l, r)$ $(1 \leq l \leq r \leq n)$, count the number of pairs $(x, y)$ $(l \leq x \leq y \leq r)$ such that the amount of $\mathtt{0}$ equals the amount of $\mathtt{1}$ in the substring $s_xs_{x+1}...s_y$.

Output the sum of counts over all possible $(l, r)$ modulo $10^9+7$.

Input: The first line contains $t$ ($1 \leq t \leq 1000$) — the number of test cases.

Each test case contains a binary string $s$ ($1 \leq |s| \leq 2 \cdot 10^5$). It is guaranteed $s$ only contains characters $\mathtt{0}$ and $\mathtt{1}$.

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

Output: For each test case, output an integer, the answer modulo $10^9+7$.

Examples:
input:
4
0000
01010101
1100111001
11000000111
output:
0
130
147
70

Note:

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

public class Main {
    static FastReader read = new FastReader();
    static StringBuilder out = new StringBuilder();

    public static void main(String[] args) throws Exception {
        int t = read.nextInt();
        while (t-- > 0) {
            char arr[] = read.nextLine().toCharArray();
            int prefix[] = new int[arr.length + 1];

            int ans = 0;
            int n = arr.length;
            int mod = 1000_000_007;

            for(int i = 0; i < arr.length; i++) {
                prefix[i + 1] = prefix[i] + (arr[i] == '0' ? 1 : -1);
            }
            int countXTotalSize = 4 * n + 100;
            int countX[] = new int[countXTotalSize];
            for(int i = 0; i <= arr.length; i++) {
                int idx = prefix[i] + (countXTotalSize / 2);
                ans = (ans + (countX[idx] * (n - i + 1)) % mod) % mod;
                countX[idx] = ((countX[idx] + (i + 1) ))% mod;
            }

            out.append(ans).append('\n');

        }
        System.out.print(out);
    }

}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

    public FastReader(String fileName) throws IOException {
        br = new BufferedReader(new FileReader(fileName));
    }

    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;
    }
}

**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
0101
1010
0110
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Problem about GCD
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given three integers $$$l$$$, $$$r$$$, and $$$G$$$, find two integers $$$A$$$ and $$$B$$$ ($$$l \le A \le B \le r$$$) such that their greatest common divisor (GCD) equals $$$G$$$ and the distance $$$|A - B|$$$ is maximized.

If there are multiple such pairs, choose the one where $$$A$$$ is minimized. If no such pairs exist, output "-1 -1".

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

Each test case consists of a single line containing three integers $$$l, r, G$$$ ($$$1 \le l \le r \le 10^{18}$$$; $$$1 \le G \le 10^{18}$$$) — the range boundaries and the required GCD.

Output: For each test case, output two integers $$$A$$$ and $$$B$$$ — the solution to the problem, or "-1 -1" if no such pair exists.

Examples:
input:
4
4 8 2
4 8 3
4 8 4
5 7 6
output:
4 6
-1 -1
4 8
6 6

Note:

Buggy Code:
// Problem: D. Problem about GCD
// Contest: Codeforces - Educational Codeforces Round 173 (Rated for Div. 2)
// URL: https://codeforces.com/contest/2043/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define int long long
#define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f3f3f3f
using namespace std;

const int N = 2e6 + 10;
const int mod = 1e9 + 7;
mt19937_64 rnd(time(NULL));
//const int mod = 998244353;

const int dx[] = { 0, 0, 1, -1 };
const int dy[] = { 1, -1, 0, 0 };
int a[N], b[N], c[N];
int n, m, k, cnt;
// how to control to return back in a tree, just excecute it normally and return some useful information from son
// if from up to down, first calculate last digui, else (namely from down to up), first digui last calculate

// a utilized thought : use mutiset to maintain the minimum of a segment of array
int gcd(int a, int b) {
	if (!b) return a;
	else return gcd(b, a % b);
}
void solve() {
	int l, r, g;
	cin >> l >> r >> g;
	int ll = (l + g - 1) / g;
	int rr = r / g;
	if (ll == rr && ll != 1) {
		cout << -1 << ' ' << -1 << endl;
		return ;
	}
	int x = 0, y = 0;
	int m_max = -INF;
	for(int i = ll; i <= min(rr, ll + 10000); i++){
		for(int j = rr; j >= ll; j--){
			if(gcd(i, j) == 1){
				int t = j - i;
				if (t > m_max) {
					m_max = t;
					x = i, y = j;
				}
				break;

			}
		}
	}
	cout << x * g << ' ' << y * g << endl;
 	// int x = 0, y = 0;
	// for (int i = ll;  ; ++i) {
		// if (gcd(i, rr) == 1) break;
		// x += 1;
	// }
	// for (int i = rr; ; --i) {
//
	// }
	return ;
}
signed main() {
//	freopen("./in.txt", "r", stdin);
	ios;
	int test = 1;
	cin >> test;
	while (test--) {
		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
1 1000000000000000000 1000000000000000000
```