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>
template <class T>
inline void read(T &res) {
  res = 0;
  bool bo = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    bo = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  if (bo) res = ~res + 1;
}
const int N = 1e5 + 5;
int n, x;
char s[N];
void work() {
  read(n);
  read(x);
  scanf("%s", s + 1);
  int now = 0, sum = 0, ans = 0;
  for (int i = 1; i <= n; i++)
    if (s[i] == '0')
      sum++;
    else
      sum--;
  for (int i = 1; i <= n; i++) {
    if (sum == 0 && now == x) return (void)puts("-1");
    if (sum) {
      int tmp = (x - now) / sum;
      if (tmp >= 0 && 1ll * sum * tmp + now == x) ans++;
    }
    if (s[i] == '0')
      now++;
    else
      now--;
  }
  printf("%d\n", ans);
}
int main() {
  int T;
  read(T);
  while (T--) work();
  return 0;
}