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() {
  int n, i, sum = 0, flag = 0;
  long long x;
  cin >> n >> x;
  string s;
  cin >> s;
  int balN;
  int arrj[n];
  sum = 0;
  for (i = 0; i < n; i++) {
    if (sum == x) flag = 1;
    arrj[i] = sum;
    if (s[i] == '0')
      sum += 1;
    else
      sum -= 1;
  }
  balN = sum;
  if (balN == 0) {
    if (flag == 1)
      cout << "-1" << '\n';
    else
      cout << "0" << '\n';
  } else {
    int ans = 0, j;
    for (j = 0; j < n; j++) {
      if ((abs(x - arrj[j]) % abs(balN)) == 0 && ((x - arrj[j]) / balN) >= 0)
        ans++;
    }
    cout << ans << '\n';
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}