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;
const int mod = 1e9 + 7, p1 = 5807, p2 = 8573;
const int N = 21, SQN = 250;
const long double pi = acos(-1);
long long power(long long x, unsigned long long y, long long mod) {
  long long ans = 1;
  x = x % mod;
  while (y) {
    if (y & 1LL) {
      ans = (ans * x) % mod;
    }
    y >>= 1LL;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long c0 = 0, c1 = 0;
    long long t0 = 0, t1 = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0') {
        t0++;
      } else {
        t1++;
      }
    }
    bool f = 0;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (t0 == t1) {
        if ((c0 - c1) == x) {
          f = 1;
          break;
        }
      } else {
        long long y = x - (c0 - c1);
        long long z = t0 - t1;
        if ((y >= 0 && z >= 0) || (y <= 0 && z <= 0)) {
          if (y < 0) {
            y *= (-1);
          }
          if (z < 0) {
            z *= (-1);
          }
          if (y % z == 0) {
            ans++;
          }
        }
      }
      if (s[i] == '0') {
        c0++;
      } else {
        c1++;
      }
    }
    if (f) {
      ans = -1;
    }
    cout << ans << "\n";
  }
  return 0;
}