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")
using namespace std;
const int MAX = 2e5 + 5;
const long long MAX2 = 11;
const long long MOD = 1000000007;
const long long MOD2 = 1000000006;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 316;
long long tc, n, x, ans, nw, tmp;
string s;
map<int, int> cnt;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n >> x >> s;
    ans = nw = 0, cnt.clear();
    cnt[0] = 1;
    for (auto i : s) {
      if (i == '0')
        ++nw;
      else
        --nw;
      ++cnt[nw];
    }
    --cnt[nw];
    if (nw == 0) {
      if (cnt.count(x))
        cout << "-1\n";
      else
        cout << "0\n";
    } else {
      for (auto i : cnt) {
        tmp = x - i.first;
        if (tmp % nw == 0 && tmp / nw >= 0) ans += i.second;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}