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 MOD = 1000000007;
long long gcdf(long long a, long long b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return max(a, b);
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int main() {
  boost();
  long long test = 1, i, j, k, l;
  cin >> test;
  while (test--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long cnt0 = 0, cnt1 = 0, co = 0, mi = INT_MAX, ma = INT_MIN;
    unordered_map<long long, int> mpp;
    for (i = 0; i < n; i++) {
      if (s[i] == '0')
        cnt0++, co++;
      else
        cnt1++, co--;
      mpp[co]++;
      ma = max(ma, co);
      mi = min(mi, co);
    }
    if (co >= 0 && mi > x) {
      if (x != 0)
        cout << 0 << endl;
      else
        cout << 1 << endl;
      continue;
    }
    if (co <= 0 && ma < x) {
      if (x != 0)
        cout << 0 << endl;
      else
        cout << 1 << endl;
      continue;
    }
    if (co == 0 && x >= mi && x <= ma) {
      cout << -1 << endl;
      continue;
    }
    long long ans = 0;
    if (x == 0) ans++;
    if (co > 0) {
      if (x <= ma) {
        long long st = 0, flag = 0;
        while (1) {
          long long left = mi + co * st;
          long long right = ma + co * st;
          if (left <= x && right >= x)
            ans += mpp[x - st * co];
          else
            flag++;
          if (flag > 10) break;
          st++;
        }
      } else {
        long long st = (x - ma) / co;
        st -= 3;
        st = max(st, (long long)0);
        int flag = 0;
        while (1) {
          long long left = mi + co * st;
          long long right = ma + co * st;
          if (left <= x && right >= x)
            ans += mpp[x - st * co];
          else
            flag++;
          if (flag > 10) break;
          st++;
        }
      }
    } else {
      if (x >= mi) {
        long long st = 0, flag = 0;
        while (1) {
          long long left = mi + co * st;
          long long right = ma + co * st;
          if (left <= x && right >= x)
            ans += mpp[x - st * co];
          else
            flag++;
          if (flag == 10) break;
          st++;
        }
      } else {
        long long st = (x - mi) / co;
        if (st < 0) st = st * -1;
        st -= 3;
        st = max(st, (long long)0);
        int flag = 0;
        while (1) {
          long long left = mi + co * st;
          long long right = ma + co * st;
          if (left <= x && right >= x)
            ans += mpp[x - st * co];
          else
            flag++;
          if (flag > 10) break;
          st++;
        }
      }
    }
    cout << ans << endl;
  }
}