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 gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool mod(long double a, long double b) { return a / b - floor(a / b); }
int occurs(vector<int> v, int n) {
  auto it = lower_bound(((v)).begin(), ((v)).end(), (n));
  auto it1 = upper_bound(((v)).begin(), ((v)).end(), (n));
  int x = it1 - it;
  return x;
}
long double max(long double a, long double b) { return a < b ? b : a; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    int a[n];
    int c = 0;
    for (int(i) = 0; (i) < (n); ++(i)) {
      if (s[i] == '0')
        c++;
      else
        c--;
      a[i] = c;
    }
    sort(a, a + n);
    int ans = 0;
    if (c == 0) {
      for (int(i) = 0; (i) < (n); ++(i)) {
        if (a[i] == x) {
          ans = -1;
          break;
        }
      }
      if (x == 0 && ans == 0) ans = 1;
      cout << ans << '\n';
    } else {
      if ((x >= 0 && c > 0) || (x <= 0 && c < 0)) {
        if (c > 0) {
          for (int(i) = 0; (i) < (n); ++(i)) {
            if ((x - a[i]) % c == 0 && (x - a[i]) >= 0) ans++;
          }
        } else {
          for (int(i) = 0; (i) < (n); ++(i)) {
            if ((x - a[i]) % c == 0 && (x - a[i]) <= 0) ans++;
          }
        }
        if (x == 0) ans++;
        cout << ans << '\n';
      } else {
        if (c > 0) {
          for (int(i) = 0; (i) < (n); ++(i)) {
            if ((x - a[i]) % c == 0 && (x - a[i]) >= 0) ans++;
          }
        } else {
          for (int(i) = 0; (i) < (n); ++(i)) {
            if ((x - a[i]) % c == 0 && (x - a[i]) <= 0) ans++;
          }
        }
        if (x == 0) ans++;
        cout << ans << '\n';
      }
    }
  }
  return 0;
}