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;
vector<int> pri;
vector<int> path;
unsigned long long power(unsigned long long x, int y, int p) {
  unsigned long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long modInverse(unsigned long long n, int p) {
  return power(n, p - 2, p);
}
unsigned long long nCrModPFermat(unsigned long long n, int r, int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long pp(long long n, long long mod) {
  if (n == 0) return 1;
  long long ans1 = pp(n / 2, mod);
  ans1 = (ans1 * ans1) % mod;
  if (n % 2) ans1 = (ans1 * 2) % mod;
  return ans1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    string s = "";
    cin >> s;
    unordered_map<int, int> mp;
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += s[i] - '0';
      mp[i + 1 - (sum)-sum]++;
    }
    sum -= (n - sum);
    if (sum == 0) {
      if (x > n || x < -n) {
        cout << 0 << endl;
        continue;
      }
      if (mp.count(x)) {
        cout << -1 << endl;
        continue;
      }
      cout << 0 << endl;
      continue;
    }
    long long ans = 0;
    if (x == 0) ans++;
    if (sum > 0) {
      if (x < -1 * n) {
        int kk = -n - x;
        x = -n + sum - kk % sum - sum;
      }
      while (x <= n) {
        ans += mp[x];
        x += sum;
      }
    } else {
      if (x > n) {
        int ll = -1 * sum;
        int kk = x - n;
        x = n - (ll - kk % ll) + ll;
      }
      while (x >= -n) {
        ans += mp[x];
        x += sum;
      }
    }
    cout << ans << endl;
  }
}