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 int logx = 30;
const long long int N = 1e6 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265;
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c >= mod) c -= mod;
  return c;
}
template <typename T>
T power(T x, T y, long long int m) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int dif[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, x;
    string second;
    cin >> n >> x >> second;
    string t = second;
    t += second;
    long long int m = 2 * n, zero = 0, one = 0;
    for (long long int i = 0; i < m; i++) {
      zero += (t[i] == '0');
      one += (t[i] == '1');
      dif[i] = zero - one;
    }
    long long int inc = dif[n] - dif[0], res = 0;
    if (inc == 0) {
      for (long long int i = 0; i < n; i++) {
        res += (dif[i] == x);
      }
      if (res) {
        res = -1;
      }
      cout << res << endl;
      continue;
    }
    for (long long int i = 0; i < n; i++) {
      long long int check = (x - dif[i]) % inc;
      long long int itr = (x - dif[i]) / inc;
      if (check || itr < 0) {
        continue;
      }
      res++;
    }
    if (x == 0) {
      res++;
    }
    cout << res << endl;
  }
}