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 cline() { cout << '\n'; }
template <typename T, typename... V>
void cline(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cline(v...);
}
void cspc() { cout << ' '; }
template <typename T, typename... V>
void cspc(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cspc(v...);
}
const long long N = 1e6;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long dif[n + 1];
    long long net = 0;
    bool flag = 0;
    if (x == 0) {
      flag = 1;
    }
    for (long long i = 0; i < n; i++) {
      if (s[i] == '0') {
        dif[i] = net + 1;
        net++;
      } else {
        dif[i] = net - 1;
        net--;
      }
      if (dif[i] == x) {
        flag = 1;
      }
    }
    if ((net == 0) && (flag)) {
      cspc(-1);
    } else if ((net == 0) && (!flag)) {
      cspc(0);
    } else {
      long long ans = 0;
      if (net < 0) {
        for (long long i = 0; i < n; i++) {
          if (((dif[i] - x) % net == 0) && ((dif[i] - x >= 0))) ans++;
        }
      } else {
        for (long long i = 0; i < n; i++) {
          if (((x - dif[i]) % net == 0) && ((dif[i] - x <= 0))) ans++;
        }
      }
      if (x == 0) {
        ans++;
      }
      cspc(ans);
    }
    cout << '\n';
  }
  return 0;
}