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 sorgu_s;
int basamak_s;
int hedef;
int su_an;
int deger;
int cevap;
int en_fazla, en_az;
int sayi[100007];
int main() {
  scanf("%d", &sorgu_s);
  for (int j = 0; j < sorgu_s; j++) {
    scanf(" %d %d ", &basamak_s, &hedef);
    su_an = 0;
    cevap = 0;
    en_fazla = -1;
    en_az = 1;
    deger = 0;
    for (int i = 0; i < basamak_s; i++) {
      scanf("%1d", &sayi[i]);
      if (sayi[i] == 0) {
        deger++;
        if (deger > en_fazla) {
          en_fazla = deger;
        }
      } else {
        deger--;
        if (deger < en_az) {
          en_az = deger;
        }
      }
    }
    if (deger == 0) {
      if ((en_az <= hedef and hedef <= 0) or
          (en_fazla >= hedef and hedef >= 0)) {
        printf("-1\n");
      } else {
        printf("0\n");
      }
    } else {
      for (int i = 0; i < basamak_s; i++) {
        if ((hedef - su_an) % deger == 0 and (hedef - su_an) / deger >= 0) {
          cevap++;
        }
        if (sayi[i] == 0) {
          su_an++;
        } else {
          su_an--;
        }
      }
      printf("%d\n", cevap);
    }
  }
}