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 INF = 1e6;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
int T, n, k, s[100005];
char c[100005];
int main() {
  for (read(T); T--;) {
    read(n), read(k), scanf("%s", c + 1), s[0] = 0;
    long long ans = 0;
    char tag = 0;
    for (int i = 1; i <= n; i++) s[i] = s[i - 1] + (c[i] == '0' ? 1 : -1);
    for (int i = 0; i <= n; i++) {
      if (s[i] == k) ans++;
      if (s[i] == k && !s[n]) tag = 1;
      if (i && s[i] < k && s[n] > 0 && (k - s[i]) % s[n] == 0) ans++;
      if (i && s[i] > k && s[n] < 0 && (s[i] - k) % (-s[n]) == 0) ans++;
    }
    if (tag)
      puts("-1");
    else
      printf("%d\n", ans);
  }
  return 0;
}