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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    int a[100001] = {0};
    int b[100001] = {0};
    int f = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == '0') {
        a[i + 1] = a[i] + 1;
        b[i + 1] = b[i];
      } else {
        a[i + 1] = a[i];
        b[i + 1] = b[i] + 1;
      }
    }
    long long ans = 0;
    int c0 = 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] - b[i] == 0) {
        c0++;
      }
      int g = a[i] - b[i];
      long double h;
      h = (x - g);
      h *= 1.0;
      if (a[n] - b[n] == 0) {
        if (g == x) ans++;
        continue;
      }
      h /= (long double)(a[n] - b[n]);
      if (h < 0) continue;
      if (ceil(h) == floor(h)) {
        ans++;
      }
    }
    if (a[n] - b[n] == 0) {
      for (int i = 1; i <= n; i++)
        if (a[i] - b[i] == x) {
          cout << -1 << '\n';
          f = 1;
          break;
        }
    }
    if (x == 0) ans++;
    if (f) continue;
    cout << ans << '\n';
  }
}