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 + 45;
long long n, x;
string s;
void solve() {
  cin >> n >> x;
  cin >> s;
  long long bal = 0;
  long long br = 0;
  map<int, int> mapa;
  if (bal == x) {
    br++;
  }
  for (auto f : s) {
    bal += (f == '0');
    bal -= (f == '1');
    mapa[bal]++;
    if (bal == x) {
      br++;
    }
  }
  if (bal == 0) {
    if (br == 0) {
      cout << 0 << endl;
    } else {
      cout << -1 << endl;
    }
    return;
  }
  for (auto f : mapa) {
    long long a = x - f.first;
    if (a % bal != 0) {
      continue;
    }
    if (a / bal >= 1) {
      br += f.second;
    }
  }
  cout << br << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}