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 int maxn = 2e5 + 5;
const long long mod = 998244353;
int mp[maxn], a[maxn];
char s[maxn];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, x;
    scanf("%d", &n);
    scanf("%d", &x);
    scanf("%s", s + 1);
    int mn = 1e9, mx = -1e9;
    for (int i = (1); i <= (n); i++) {
      if (s[i] == '0')
        a[i] = a[i - 1] + 1;
      else
        a[i] = a[i - 1] - 1;
      mx = max(mx, a[i]), mn = min(mn, a[i]);
    }
    for (int i = (0); i <= (mx - mn); i++) mp[i] = 0;
    for (int i = (1); i <= (n); i++) mp[a[i] - mn]++;
    if (a[n] == 0) {
      if (x < mn || x > mx || mp[x - mn] == 0) {
        if (x != 0)
          puts("0");
        else
          puts("1");
      } else
        puts("-1");
      continue;
    }
    if (a[n] > 0) {
      int l = 0, r = 1e9, pos = -1;
      while (l <= r) {
        int mid = l + r >> 1;
        if (x >= 1ll * mid * a[n] + mn && x <= 1ll * mid * a[n] + mx)
          pos = mid, r = mid - 1;
        else if (x < 1ll * mid * a[n] + mn)
          r = mid - 1;
        else
          l = mid + 1;
      }
      long long ans = 0;
      if (pos == -1)
        ans = 0;
      else {
        int mpx = -1;
        for (int i = pos * a[n] + mn; i <= pos * a[n] + mx; i++) {
          if (i <= x)
            mpx++;
          else
            break;
        }
        for (int i = mpx; i >= 0; i -= a[n]) ans += mp[i];
      }
      if (x == 0) ans++;
      printf("%lld", ans);
      puts("");
    } else {
      int l = 0, r = 1e9, pos = -1;
      while (l <= r) {
        int mid = l + r >> 1;
        if (x >= 1ll * mid * a[n] + mn && x <= 1ll * mid * a[n] + mx)
          pos = mid, r = mid - 1;
        else if (x < 1ll * mid * a[n] + mn)
          l = mid + 1;
        else
          r = mid - 1;
      }
      long long ans = 0;
      if (pos == -1)
        ans = 0;
      else {
        int mpx = -1;
        for (int i = pos * a[n] + mn; i <= pos * a[n] + mx; i++) {
          if (i <= x)
            mpx++;
          else
            break;
        }
        for (int i = mpx; i <= mx - mn; i += (-a[n])) ans += mp[i];
      }
      if (x == 0) ans++;
      printf("%lld", ans);
      puts("");
    }
  }
}