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;
template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const double EPS = 1e-10;
const double PI = acos(-1);
class BInfinitePrefixes {
 public:
  void solve(std::istream& in, std::ostream& out) {
    ios_base::sync_with_stdio(false);
    in.tie(nullptr), out.tie(nullptr);
    int tt;
    in >> tt;
    while (tt--) {
      int N, X;
      in >> N >> X;
      string S;
      in >> S;
      int bal = 0;
      map<int, int> mp;
      for (auto c : S) {
        if (c == '0')
          ++bal;
        else
          --bal;
        mp[bal]++;
      }
      if (bal == 0) {
        if (mp.count(X))
          out << "-1\n";
        else
          out << "0\n";
      } else if (bal > 0) {
        int ans = 0;
        for (auto p : mp) {
          if (X >= p.first and
              (X % bal + bal) % bal == (p.first % bal + bal) % bal)
            ans += p.second;
        }
        if (X == 0) ++ans;
        out << ans << '\n';
      } else {
        int ans = 0;
        X *= -1;
        bal *= -1;
        for (auto p : mp) {
          int pf = p.first * (-1);
          if (X >= pf and (X % bal + bal) % bal == (pf % bal + bal) % bal)
            ans += p.second;
        }
        if (X == 0) ++ans;
        out << ans << '\n';
      }
    }
  }
};
int main() {
  BInfinitePrefixes solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}