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 N = 1e5 + 5;
long long a[N];
long long bal[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    long long x;
    cin >> n >> x;
    for (int i = 1; i <= n; i++) {
      char c;
      cin >> c;
      a[i] = c - '0';
    }
    int zeros = 0, ones = 0;
    for (int i = 0; i <= n; i++) {
      if (i) {
        if (a[i])
          ones++;
        else
          zeros++;
      }
      bal[i] = zeros - ones;
    }
    long long tot = bal[n];
    long long ans = 0;
    if (x == 0) ans++;
    for (int i = 1; i <= n; i++) {
      if (x == bal[i] && tot == 0) {
        ans = -1;
        break;
      }
      if (tot == 0) {
        continue;
      }
      if ((x - bal[i]) % tot == 0) {
        if ((x - bal[i]) / tot >= 0) ans++;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}