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 ll = long long;
using namespace std;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  int t;
  cin >> t;
  for (ll _ = 0, _Len = (t); _ < _Len; ++_) {
    ll n, x;
    string bit_s;
    cin >> n >> x >> bit_s;
    vector<ll> acc(n);
    map<ll, ll> cnt;
    acc[0] = (bit_s[0] == '0' ? 1 : -1);
    cnt[acc[0]]++;
    for (ll i = 1, iLen = (n); i < iLen; ++i) {
      acc[i] = acc[i - 1] + (bit_s[i] == '0' ? 1 : -1);
      cnt[acc[i]]++;
    }
    if (acc[n - 1] == 0) {
      cout << (cnt[x] > 0 ? -1 : (x == 0 ? 1 : 0)) << "\n";
      continue;
    }
    ll ans = (x == 0 ? 1 : 0), m = acc[n - 1];
    for (auto &[c, v] : cnt) {
      if ((x - c) % m == 0 && (x - c) / m >= 0) ans += v;
    }
    cout << ans << "\n";
  }
}