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;
using LL = long long;
const int mo = 1e9 + 7;
LL qpow(LL a, LL b) {
  LL ans = 1;
  a %= mo;
  while (b) {
    if (b & 1) ans = ans * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return ans;
}
const int maxn = 1e5 + 5;
char s[maxn];
int sum[maxn];
map<int, int> my;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, x;
    my.clear();
    scanf("%d%d", &n, &x);
    int tmp = x;
    scanf("%s", s + 1);
    int num0 = 0, num1 = 0;
    int i;
    for (i = 1; s[i]; i++) {
      if (s[i] == '0') {
        num0++;
        sum[i] = sum[i - 1] + 1;
      } else {
        num1++;
        sum[i] = sum[i - 1] - 1;
      }
      my[sum[i]]++;
    }
    int delt;
    LL ans = 0;
    if (num1 == num0) {
      if (my[x]) {
        printf("-1\n");
      } else
        printf("0\n");
      continue;
    }
    delt = num0 - num1;
    if (x > n) {
      if (delt < 0) {
        printf("0\n");
        continue;
      } else {
        x -= ((x - n) / delt) * delt;
        if (x > n) x -= delt;
      }
    }
    if (x < -n) {
      if (delt > 0) {
        printf("0\n");
        continue;
      } else {
        x += ((-n - x) / (-delt)) * (-delt);
        if (x < -n) x -= delt;
      }
    }
    while (x <= n && x >= -n) {
      ans += my[x];
      x -= delt;
    }
    if (tmp == 0) ans++;
    printf("%lld\n", ans);
  }
  return 0;
}