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 GÑC optimize("Pivo na polu")
using namespace std;
const int mod = 1e9 + 7;
const int INF32 = 2009000999;
const int SZ = 2e5 + 11;
const int logSZ = 20;
const long long INF64 = 4e18;
const long double eps = 1e-12;
void Solve() {
  int n, x, curr = 0, ans = 0;
  string s;
  cin >> n >> x >> s;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    curr += (s[i] == '0' ? 1 : -1);
    mp[curr]++;
  }
  mp[curr]--;
  mp[0]++;
  if (!curr) {
    cout << (mp[x] ? -1 : 0) << endl;
    return;
  }
  for (auto j : mp) {
    int k = x - j.first;
    if (k % curr || k / curr < 0) continue;
    ans += j.second;
  }
  cout << ans << '\n';
}
signed main() {
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int q = 1;
  cin >> q;
  while (q--) Solve();
}