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;
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B, T M) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M) {
  return BigMod(A, M - 2, M);
}
long long gcdr(long long a, long long b) {
  if (a == 0) return b;
  return gcdr(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void FastIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
}
const long long size3 = 300005;
const long long size2 = 200005;
const long long size1 = 100005;
int main() {
  FastIO();
  long long t, i, j, m, x, y, n;
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> m;
    string s;
    cin >> s;
    long long one = 0, zero = 0;
    for (i = 0; i < n; i++) {
      if (s[i] == '1') {
        one++;
      } else {
        zero++;
      }
    }
    if (m == 0 && one == zero) {
      cout << -1 << endl;
      continue;
    }
    if ((one == zero)) {
      long long ans = 0;
      long long cOne = 0, cZero = 0;
      ans += (m == (cZero - cOne));
      for (i = 0; i < 10 * n; i++) {
        if (s[i % n] == '1') {
          cOne++;
        } else {
          cZero++;
        }
        ans += (m == (cZero - cOne));
      }
      if (ans)
        cout << -1 << endl;
      else {
        cout << 0 << endl;
      }
      continue;
    }
    if (!zero) {
      if (m > 0) {
        cout << 0 << endl;
        continue;
      } else {
        cout << 1 << endl;
        continue;
      }
    }
    if (!one) {
      if (m < 0) {
        cout << 0 << endl;
        continue;
      } else {
        cout << 1 << endl;
        continue;
      }
    }
    long long tZ = 0, tO = 0;
    long long ans = 0;
    ans += (m == 0);
    long long diff = 0;
    diff = zero - one;
    for (i = 0; i < n; i++) {
      if (s[i] == '1') {
        tO++;
      } else {
        tZ++;
      }
      long long first = tZ - tO;
      long long rem = m - first;
      if (rem == 0) {
        ans++;
      } else if ((rem < 0 && diff > 0) || (rem > 0 && diff < 0)) {
      } else {
        if ((rem % diff) == 0) {
          ans++;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}