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;
void INP();
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  if (0) return;
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  if (0) return;
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void alpha() {
  long long int n, x;
  cin >> n >> x;
  string str;
  cin >> str;
  long long int arr[n];
  map<long long int, long long int> mp;
  long long int cnt_0 = 0, cnt_1 = 0;
  for (long long int i = 0; i < n; i++) {
    if (str[i] == '1')
      cnt_1++;
    else
      cnt_0++;
    arr[i] = cnt_0 - cnt_1;
    mp[arr[i]]++;
  }
  long long int diff = cnt_0 - cnt_1;
  long long int ans = 0;
  if (diff == 0 and mp[x] != 0) {
    cout << -1 << '\n';
    return;
  } else if (diff == 0 and mp[x] == 0) {
    cout << 0 << '\n';
    return;
  }
  if (x == 0) ans++;
  for (long long int i = 0; i < n; i++) {
    if ((x - arr[i]) >= 0 and diff > 0 and (x - arr[i]) % diff == 0)
      ans++;
    else if ((x - arr[i]) <= 0 and diff < 0 and
             abs(x - arr[i]) % abs(diff) == 0)
      ans++;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  INP();
  long long int t = 1;
  cin >> t;
  while (t--) alpha();
  return 0;
}
inline void INP() {}