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;
inline long long read() {
  register long long t = 0, f = 0;
  register char v = getchar();
  while (v < '0') f |= (v == '-'), v = getchar();
  while (v >= '0') {
    t = (t << 3) + (t << 1) + v - 48;
    v = getchar();
  }
  return f ? -t : t;
}
long long n, t, x, ans, mx, mn;
char s[100002];
long long mp[200002];
long long cnt0, cnt1;
signed main() {
  t = read();
  while (t--) {
    n = read();
    mx = 0;
    mn = 0;
    x = read();
    memset(mp, 0, sizeof(mp));
    cnt0 = cnt1 = ans = 0;
    scanf("%s", s + 1);
    for (register long long i = 1; i <= n; ++i) {
      if (s[i] == '0')
        ++cnt0;
      else
        ++cnt1;
    }
    long long tmp = cnt0 - cnt1;
    cnt0 = cnt1 = 0;
    if (tmp == 0) {
      bool ia = 0;
      for (register long long i = 1; i <= n; ++i) {
        if (s[i] == '0')
          ++cnt0;
        else
          ++cnt1;
        if (cnt0 - cnt1 == x) {
          puts("-1");
          ia = 1;
          break;
        }
      }
      if (!ia) puts("0");
      continue;
    }
    if (x == 0) ++ans;
    for (register long long i = 1; i <= n; ++i) {
      if (s[i] == '0')
        ++cnt0;
      else
        ++cnt1;
      if ((x - cnt0 + cnt1) * tmp >= 0) {
        if ((x - cnt0 + cnt1) % tmp == 0) ++ans;
      }
    }
    printf("%lld\n", ans);
  }
}