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("Ofast,no-stack-protector")
#pragma GCC optimize("unroll-loops")
using namespace std;
void my_file_in_out() {
  freopen("inp.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
inline void OK(string t = "") { cerr << "OK " << t << '\n'; }
inline long double time() { return clock() / (long double)CLOCKS_PER_SEC; }
inline void stop(string t = "PROCESS FINISHED") {
  cerr << t << '\n';
  exit(0);
}
void solve() {
  long long n, x;
  string a;
  cin >> n >> x >> a;
  vector<long long> c(2, 0);
  map<long long, vector<int>> d;
  d[0].push_back(0);
  for (int i = 0; i < n; i++) {
    c[a[i] - '0']++;
    d[c[0] - c[1]].push_back(i + 1);
  }
  if ((c[0] == c[1])) {
    if (d.find(x) != d.end())
      cout << -1 << '\n';
    else
      cout << 0 << '\n';
    return;
  }
  long long p = c[0] - c[1];
  vector<long long> ans;
  for (auto &el : d) {
    long long dt = x - el.first;
    long long cd = dt / p;
    if (cd >= 0 && cd * p == dt)
      for (auto e : el.second) ans.push_back(cd * n + e);
  }
  sort(ans.begin(), ans.end());
  ans.erase(unique(ans.begin(), ans.end()), ans.end());
  cout << ((int)ans.size()) << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}