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() {
  map<long long, long long> mp;
  int TC;
  cin >> TC;
  while (TC--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long zero = 0, ones = 0;
    mp[0] = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] == '0')
        zero++;
      else
        ones++;
      mp[i + 1] = zero - ones;
    }
    zero = zero - ones;
    long long res = 0;
    double k;
    if (zero) {
      for (long long i = 0; i < n; i++) {
        k = (double)(x - mp[i]) / (double)zero;
        if (!(k - floor(k)) && k >= 0) {
          res++;
        }
      }
      cout << res << endl;
    } else {
      for (long long i = 0; i <= n; i++) {
        if (mp[i] == x) res++;
      }
      if (res)
        cout << -1 << endl;
      else
        cout << 0 << endl;
    }
  }
}