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;
int sign(long long x) {
  if (x < 0) return -1;
  if (x > 0) return 1;
  return 0;
}
int solve(int N, long long X, const string& S) {
  vector<int> PS(N + 1);
  for (int i = (1), _b = (N); i <= _b; ++i) {
    PS[i] = PS[i - 1] + (S[i - 1] == '0' ? +1 : -1);
  }
  if (PS[N] == 0) {
    for (int i = (1), _b = (N); i <= _b; ++i)
      if (PS[i] == X) return -1;
    return 0;
  }
  int res = 0;
  for (int i = (1), _b = (N); i <= _b; ++i) {
    long long dif = X - PS[i];
    if ((dif % PS[N]) == 0) {
      if (dif == 0 || sign(dif) == sign(PS[N])) ++res;
    }
  }
  if (X == 0) ++res;
  return res;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int TC;
  cin >> TC;
  for (int tc = (1), _b = (TC); tc <= _b; ++tc) {
    int N, X;
    cin >> N >> X;
    string S;
    cin >> S;
    int res = solve(N, X, S);
    cout << res << '\n';
  }
  return 0;
}