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 = 100 * 1000 + 10;
int t, n, x, dp[maxn], mn, mx, ans;
string s;
int main() {
  cin >> t;
  for (int z = 0; z < t; z++) {
    mn = mx = ans = 0;
    cin >> n >> x >> s;
    for (int i = 0; i < n; i++) {
      dp[i + 1] = dp[i];
      if (s[i] == '0')
        dp[i + 1]++;
      else
        dp[i + 1]--;
      mn = min(mn, dp[i + 1]);
      mx = max(mx, dp[i + 1]);
    }
    if (dp[n] == 0) {
      if (mn <= x && x <= mx)
        cout << -1 << endl;
      else
        cout << 0 << endl;
    } else {
      for (int i = 0; i < n; i++) {
        if ((x - dp[i]) % dp[n] == 0 && (long long)(x - dp[i]) * dp[n] >= 0)
          ans++;
      }
      cout << ans << endl;
    }
  }
  return 0;
}