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, long long int x, string s) {
  long long int a[100010], i, d, ans;
  a[0] = 0;
  d = 0;
  for (i = 0; i < n; i++)
    if (s[i] == '0') {
      a[i + 1] = a[i] + 1;
      d++;
    } else {
      a[i + 1] = a[i] - 1;
      d--;
    }
  if (d == 0) {
    for (i = 0; i < n + 1; i++)
      if (a[i] == x) {
        cout << -1;
        return;
      }
    cout << 0;
    return;
  }
  ans = 0;
  for (i = 0; i < n; i++) {
    a[i] = x - a[i];
    if (a[i] % d == 0) {
      a[i] = a[i] / d;
      if (a[i] >= 0) ans++;
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int q, i, nn, xx;
  string ss;
  cin >> q;
  for (i = 1; i < q + 1; i++) {
    cin >> nn >> xx >> ss;
    solve(nn, xx, ss);
    cout << "\n";
  }
}