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 mod = (1 ? 1e9 + 7 : 998244353);
const int N = 2e5 + 2;
const int M = 2e7 + 2;
const int inf = 2e9;
const long long linf = 1e18;
int f(char s) {
  if (s == '0') return 1;
  return -1;
}
void solve() {
  int n, x;
  string s;
  cin >> n >> x >> s;
  vector<int> p(n, 0);
  p[0] = f(s[0]);
  for (int i = 1; i < n; i++) p[i] = p[i - 1] + f(s[i]);
  if (!p[n - 1]) {
    int ans = 0;
    for (int i = 0; i < n; i++)
      if (p[i] == x) ans = -1;
    cout << ans << '\n';
    return;
  }
  int ans = !x;
  for (int i = 0; i < n; i++) {
    if (p[n - 1] > 0 && (x - p[i]) % p[n - 1] == 0 && x - p[i] >= 0) ans++;
    if (p[n - 1] < 0 && (p[i] - x) % p[n - 1] == 0 && p[i] - x >= 0) ans++;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}