Problem: You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...

Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.

A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers n and x (1 ≤ n ≤ 10^5, -10^9 ≤ x ≤ 10^9) — the length of string s and the desired balance, respectively.

The second line contains the binary string s (|s| = n, s_i ∈ \{0, 1\}).

It's guaranteed that the total sum of n doesn't exceed 10^5.

Output

Print T integers — one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.

Example

Input


4
6 10
010010
5 3
10101
1 0
0
2 0
01


Output


3
0
1
-1

Note

In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int T = 1;
  scanf("%d ", &T);
  while (T--) {
    int n, i, j;
    long long int x;
    scanf("%d %I64d ", &n, &x);
    char s[n];
    for (i = 0; i < n; i++) {
      scanf("%c", &s[i]);
    }
    vector<long long int> L;
    long long int sum = 0, Max = 0, Min = 0;
    for (i = 0; i < n; i++) {
      if (s[i] == '0')
        sum++;
      else
        sum--;
      L.push_back(sum);
      Max = max(Max, sum);
      Min = min(Min, sum);
    }
    sort(L.begin(), L.end());
    if (sum == 0) {
      bool found = false;
      for (i = 0; i < n; i++) {
        if (L[i] == x) {
          found = true;
          break;
        }
      }
      if (found) {
        printf("-1");
      } else {
        printf("0");
      }
    } else if (sum > 0) {
      long long int ans = (x == 0);
      for (i = 0; i < n; i++) {
        if ((x - L[i]) % sum == 0 && (x - L[i]) / sum >= 0) ans++;
      }
      printf("%I64d", ans);
    } else {
      long long int ans = (x == 0);
      long long int s = -sum;
      for (i = 0; i < n; i++) {
        long long int current = -(x - L[i]);
        if (current % s == 0 && current / s >= 0) ans++;
      }
      printf("%I64d", ans);
    }
    printf("\n");
  }
  return 0;
}