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>
#pragma GCC optimize("O3")
using namespace std;
string itos(int n) {
  stringstream ss;
  ss << n;
  string second = ss.str();
  return second;
}
long long ppow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long x;
    cin >> x;
    string second;
    cin >> second;
    long long c0 = 0, c1 = 0;
    for (char c : second) {
      if (c == '1') {
        c1++;
      } else
        c0++;
    }
    long long c = 0;
    long long ans = 0;
    if (x == 0) {
      ans = 1;
    }
    if (c0 == c1 && x == 0) {
      cout << "-1\n";
      continue;
    }
    for (int i = 0; i < n; ++i) {
      if (second[i] == '0') {
        c++;
      } else {
        c--;
      }
      if (c == x) {
        ans++;
      }
    }
    c = c0 - c1;
    bool ok = 1;
    for (int i = 0; i < n; ++i) {
      if (second[i] == '0') {
        c++;
      } else {
        c--;
      }
      long long need = x - c;
      if (need == 0 && c0 - c1 == 0) {
        ok = 0;
        break;
      }
      if (need == 0) {
        ans++;
      } else if ((c0 - c1 > 0 && need > 0) || (c0 - c1 < 0 && need < 0)) {
        need = abs(need);
        long long cur = abs(c0 - c1);
        if (need % cur == 0) {
          ans++;
        }
      }
    }
    if (!ok) {
      cout << "-1\n";
    } else
      cout << ans << "\n";
  }
}