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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    long long n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<long long> m;
    unordered_map<long long, long long> st;
    long long a = 0, b = 0;
    long long l = 100000000000, r = 0;
    for (int i = (0); i < (n); i++) {
      if (s[i] == '0')
        a++;
      else
        b++;
      st[a - b]++;
      l = min(l, a - b);
      r = max(r, a - b);
    }
    long long df = a - b;
    if (df * k < 0 && st[k] == 0)
      cout << 0 << endl;
    else if (df == 0) {
      if (st[k] == 0)
        cout << 0 << endl;
      else
        cout << -1 << endl;
    } else {
      long long i;
      if (df > 0)
        i = abs(k - r) / abs(df);
      else
        i = abs(k - l) / abs(df);
      if (k >= l && k <= r) i = 0;
      i -= 5;
      if (i < 0) i = 0;
      long long df1 = df, k1 = k, l1 = l, r1 = r;
      long long ans = 0;
      if (k == 0) ans++;
      while (1) {
        df1 = i * df;
        k1 = k - df1;
        ans += st[k1];
        l1 = l + df1;
        r1 = r + df1;
        if ((df > 0 && k < l1) || (df < 0 && k > r1)) break;
        i++;
      }
      cout << ans << endl;
    }
  }
}