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 long long maxx = 1e9 + 7;
bool mysort(const long long &b, const pair<long long, long long> &a) {
  return a.first > b;
}
bool sortbyboth(const pair<long long, long long> &a,
                const pair<long long, long long> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
long long inline power(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
long long inv(long long n, long long p) { return power(n, p - 2, p); }
inline vector<long long> allinv(long long n, long long p) {
  vector<long long> arr(n);
  arr[1] = 1;
  for (long long i = (2); i < (n); i += 1) {
    arr[i] = (p - (p / i) * arr[p % i] % p) % p;
  }
  return arr;
}
vector<long long> inline allncr(long long n, long long p) {
  vector<long long> nci(n + 1);
  nci[0] = 1;
  nci[n] = 1;
  vector<long long> v = allinv(n, p);
  for (long long i = (1); i < (n); i += 1) {
    nci[i] = (nci[i - 1] * (n - i - 1)) % p;
  }
  return nci;
}
long long max3(long long a, long long b, long long c) {
  return max(max(a, b), c);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool inline isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void prt(int i) { i ? cout << "YES\n" : cout << "NO\n"; }
long long getSum(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum = sum + n % 10;
    n = n / 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long a = 0, b = 0;
    vector<long long> cnt(n);
    for (long long i = 0; i < n; i++) {
      s[i] == '0' ? a++ : b++;
      cnt[i] = (a - b);
    }
    long long diff = a - b;
    long long ans = 0;
    if (diff == 0) {
      for (long long i = 0; i < n; i++) {
        if (x == cnt[i]) ans++;
      }
      if (ans == 0) {
        if (x == 0)
          cout << 1 << "\n";
        else
          cout << 0 << "\n";
      } else
        cout << -1 << "\n";
    } else {
      for (long long i = 0; i < n; i++) {
        if ((abs(x - cnt[i])) % abs(diff) == 0) {
          if ((x - cnt[i]) / diff >= 0) ans++;
        }
      }
      if (x != 0)
        cout << ans << "\n";
      else
        cout << ans + 1 << "\n";
    }
  }
}