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;
void fs(long long int &number) {
  bool negative = false;
  register long long int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long int power(long long int a, long long int x) {
  a %= 1000000007;
  long long int res = 1;
  while (x > 0) {
    if (x & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    x >>= 1;
  }
  return res;
}
long long int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long int min(long long int a, long long int b) {
  if (a > b) {
    return b;
  } else {
    return a;
  }
}
long long int madd(long long int a, long long int b) {
  long long int ans;
  ans = (a % 1000000007 + b % 1000000007) % 1000000007;
  return ans;
}
long long int mmult(long long int a, long long int b) {
  long long int ans = 1;
  ans = ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  return ans % 1000000007;
}
long long int modinverse(long long int a) { return power(a, 1000000007 - 2); }
void solve() {
  long long int n, x;
  cin >> n >> x;
  long long int c1 = 0, c0 = 0;
  string s;
  cin >> s;
  vector<long long int> bal;
  bal.push_back(0);
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '0') {
      c0++;
    } else {
      c1++;
    }
    if (i != n - 1) {
      bal.push_back(c0 - c1);
    }
  }
  long long int temp = c0 - c1;
  long long int count = 0, flag = 0;
  for (long long int i = 0; i < bal.size(); i++) {
    long long int z = x - bal[i];
    if (z == 0) {
      if (temp == 0) {
        flag = 1;
        break;
      }
      count++;
      continue;
    }
    if (temp != 0) {
      if (z % temp == 0) {
        long long int y = z / temp;
        if (y > 0) {
          count++;
        }
      }
    }
  }
  if (flag) {
    cout << -1;
    cout << endl;
  } else {
    cout << count;
    cout << endl;
  }
}
signed main() {
  long long int test;
  cin >> test;
  for (long long int i = 1; i <= test; i++) {
    solve();
  }
}