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;
void solve() {
  long long n, x;
  cin >> n >> x;
  string s;
  cin >> s;
  vector<long long> v(s.size());
  v[0] = (s[0] == '0') ? 1 : -1;
  for (long long i = 1; i < s.size(); i++)
    v[i] = v[i - 1] + ((s[i] == '0') ? 1 : -1);
  const long long p = 1e5 + 5;
  vector<long long> bal(2 * n + 1);
  for (long long i = 0; i <= 2 * n; i++) bal[i] = 0;
  for (long long i = 0; i < s.size(); i++) bal[v[i] + n]++;
  long long res = 0;
  for (long long i = 0; i <= 2 * n; i++) {
    if (bal[i] == 0) continue;
    if (v.back() == 0 && i - n == x) {
      cout << "-1\n";
      return;
    }
    if (v.back() == 0) continue;
    if (abs(x + n - i) % abs(v.back()) != 0) continue;
    if ((x + n - i) * v.back() < 0) continue;
    res += bal[i];
  }
  cout << (res + (x == 0)) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  for (; t; t--) solve();
}