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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long MAXN = 1123456;
const long long N = 2e6;
const long long inf = 3e18;
mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void vout(T s) {
  cout << s << endl;
  exit(0);
}
void main_solve() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    long long delta = 0;
    map<long long, long long> mp;
    long long sc = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i - 1] == '0')
        sc++;
      else
        sc--;
      mp[sc]++;
    }
    if (sc == 0) {
      if (mp[x])
        cout << "-1\n";
      else
        cout << (x == 0) << "\n";
    } else {
      long long t = abs(sc);
      long long ans = 0;
      for (auto i : mp) {
        if (sc > 0) {
          if (i.first <= x) {
            if ((x - i.first) % t == 0) ans += i.second;
          }
        } else {
          if (i.first >= x) {
            if (abs(x - i.first) % t == 0) ans += i.second;
          }
        }
      }
      cout << ans + (x == 0) << "\n";
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  main_solve();
  return 0;
}