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 MAXN = 1e5 + 15;
char S[MAXN];
long long D[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long N, X;
    cin >> N >> X >> S;
    long long C[2] = {};
    for (int i = 0; i < N; ++i) {
      ++C[S[i] - '0'];
      D[i] = C[0] - C[1];
    }
    long long x = C[0] - C[1];
    int ans = X == 0 ? 1 : 0;
    if (x == 0) {
      for (int i = 0; i < N; ++i)
        if (X - D[i] == 0) ++ans;
      if (ans > 0) ans = -1;
      cout << ans << endl;
    } else {
      for (int i = 0; i < N; ++i) {
        if (((X - D[i]) >= 0 && x > 0) || ((X - D[i]) <= 0 && x < 0)) {
          if ((X - D[i]) % x == 0) ++ans;
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}