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;
void solve() {
  long long int n, x;
  cin >> n >> x;
  string s;
  cin >> s;
  long long int arr[n];
  memset(arr, 0, sizeof(arr));
  if (s[0] == '0')
    arr[0] = 1;
  else if (s[0] == '1')
    arr[0] = -1;
  for (long long int i = 1; i < n; i++) {
    arr[i] = arr[i - 1];
    if (s[i] == '0') {
      arr[i] += 1;
    } else {
      arr[i] += -1;
    }
  }
  long long int k = arr[n - 1];
  long long int cnt = 0;
  long long int found = 0;
  if (k == 0) {
    for (long long int i = 0; i < n; i++) {
      if (arr[i] == x) {
        cout << -1;
        return;
      }
    }
  } else {
    for (long long int i = 0; i < n; i++) {
      if ((x - arr[i]) % k == 0 && (x - arr[i]) / k >= 0) {
        cnt++;
        found = 1;
      }
    }
  }
  if (x == 0) {
    cout << cnt + 1;
  } else {
    cout << cnt;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int T = 1;
  cin >> T;
  while (T--) {
    solve();
    cout << endl;
  }
  return 0;
}