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;
int takemod(int a, int mod) {
  a %= mod;
  if (a < 0) a += mod;
  return a;
}
int fast_exp(int base, int expo, int mod) {
  int res = 1;
  while (expo > 0) {
    if (expo & 1) res = (res * base) % mod;
    base = (base * base) % mod;
    expo >>= 1;
  }
  return res;
}
int modinv(int a, int mod) {
  return takemod(fast_exp(takemod(a, mod), mod - 2, mod), mod);
}
const long long N = 1e5 + 5, A = 1e5 + 10, md = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    int res = 0;
    int d = 0;
    int cur = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0')
        cur++;
      else
        cur--;
      if (cur == x) res++;
    }
    d = cur;
    if (d == 0) {
      if (res > 0)
        res = -1;
      else if (x == 0)
        res++;
      cout << res << endl;
      continue;
    }
    res = 0;
    cur = 0;
    if (x == 0) res++;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0')
        cur++;
      else
        cur--;
      if ((x - cur) % d == 0) {
        if (x - cur > 0 && d > 0) res++;
        if (x - cur < 0 && d < 0) res++;
        if (x - cur == 0) res++;
      }
    }
    cout << res << endl;
  }
}