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;
map<int, int> bin;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n, x;
    scanf("%d %d", &n, &x);
    int cir = 0;
    for (int i = 1; i <= n; i++) {
      ++bin[cir];
      int a;
      scanf("%1d", &a);
      if (!a)
        ++cir;
      else
        --cir;
    }
    long long ans = 0;
    map<int, int>::iterator iter;
    if (!cir) {
      if (!bin[x])
        puts("0");
      else
        puts("-1");
      bin.clear();
      continue;
    }
    for (iter = bin.begin(); iter != bin.end(); iter++) {
      int u = x - iter->first;
      int v = iter->second;
      if (u % cir == 0 && u / cir >= 0) ans += v;
    }
    printf("%lld\n", ans);
    bin.clear();
  }
  return 0;
}