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>
bool isT = true;
void solv(int curt);
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  if (isT) {
    int t;
    std::cin >> t;
    for (int i = 1; i <= t; i++) {
      solv(i);
    }
    return 0;
  }
  solv(0);
  return 0;
}
void solv(int curt) {
  int n, x;
  std::cin >> n >> x;
  std::string s;
  std::cin >> s;
  int z = 0;
  for (char c : s) z += c == '0';
  int other = z - (n - z), cur = 0, ans = 0;
  for (int i = 0; i < n; i++) {
    if (!other && cur == x) {
      std::cout << "-1\n";
      return;
    }
    ans += (other && !(abs(x - cur) % abs(other)) && (x - cur) / other >= 0);
    cur += s[i] == '0';
    cur -= s[i] == '1';
  }
  std::cout << ans << "\n";
}