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;
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    string a;
    cin >> a;
    long long s[n];
    long long mx = 0, mn = 0;
    for (long long i = 0; i < n; i++) s[i] = 0;
    long long bal = 0;
    for (long long i = 0; i < n; i++) {
      bal = bal + (a[i] == '0') - (a[i] == '1');
      s[i] = bal;
      mx = max(bal, mx);
      mn = min(bal, mn);
    }
    long long ats = x == 0;
    if (bal == 0) {
      if (mn <= x && x <= mx) {
        cout << "-1";
      } else
        cout << ats;
      cout << "\n";
      continue;
    }
    for (long long i = 0; i < n; i++) {
      if ((x - s[i]) % bal == 0 && (x - s[i]) / bal >= 0) {
        ats++;
      }
    }
    cout << ats << "\n";
  }
  return 0;
}