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 t, n, x;
char s[100005];
int f[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> x;
    cin >> s + 1;
    for (int i = 1; i <= n; i++) {
      f[i] = f[i - 1] + (s[i] == '0') - (s[i] == '1');
    }
    cerr << "ans: ";
    if (f[n] != 0) {
      int ans = 0;
      for (int i = 1; i <= n; i++) {
        ans += ((x - f[i]) / f[n] >= 0) && ((f[i] - x) % f[n] == 0);
      }
      cout << ans + (x == 0) << endl;
    } else {
      bool fl = false;
      for (int i = 1; i <= n; i++) {
        if (f[i] == x) {
          fl = true;
          break;
        }
      }
      if (fl) {
        cout << -1 << endl;
      } else {
        cout << 0 + (x == 0) << endl;
      }
    }
  }
  return 0;
}