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;
long long t, n, x, b[200005] = {0}, ans, ta, kk, kkk, ttt = 0;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> t;
  for (int jjj = 0; jjj < t; jjj++) {
    ta = 0;
    ans = 0;
    kk = 0;
    kkk = 0;
    cin >> n >> x;
    if (x == 0) {
      ta = 1;
    }
    for (int i = 0; i <= 2 * n; i++) {
      b[i] = 0;
    }
    cin >> s;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0') {
        ans++;
        b[ans + n]++;
      } else {
        ans--;
        b[ans + n]++;
      }
    }
    if (ans == 0) {
      if (x <= n && x >= -n) {
        if (b[x + n] != 0) {
          cout << -1 << endl;
        } else {
          cout << ta << endl;
        }
      } else {
        cout << ta << endl;
      }
    } else {
      if (ans > 0) {
        kk = (x + n) / ans;
        kkk = (x - n - 1) / ans + 1;
        if (x <= n) {
          kkk = 0;
        }
        for (int i = kkk; i <= kk + 1; i++) {
          if (x - i * ans + n >= 0 && x - i * ans + n <= 2 * n) {
            ta = ta + b[x - i * ans + n];
          }
        }
        cout << ta << endl;
      }
      if (ans < 0) {
        kk = (n + x) / ans;
        kkk = (x - n) / ans;
        for (int i = max(kk - 1, ttt); i <= kkk + 1; i++) {
          if (x - i * ans + n >= 0 && x - i * ans + n <= 2 * n) {
            ta = ta + b[x - i * ans + n];
          }
        }
        cout << ta << endl;
      }
    }
  }
}