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;
template <class T1, class T2>
istream& operator>>(istream& I, pair<T1, T2>& p) {
  return I >> p.first >> p.second;
}
template <class T1, class T2>
ostream& operator<<(ostream& O, const pair<T1, T2>& p) {
  return O << '(' << p.first << ' ' << p.second << ')';
}
template <class T>
istream& operator>>(istream& I, vector<T>& v) {
  for (T& e : v) I >> e;
  return I;
}
template <class T>
ostream& operator<<(ostream& O, const vector<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
template <class T>
ostream& operator<<(ostream& O, const unordered_set<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
template <class T>
ostream& operator<<(ostream& O, const set<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
template <class T>
ostream& operator<<(ostream& O, const unordered_multiset<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
template <class T>
ostream& operator<<(ostream& O, const multiset<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
template <class T1, class T2>
ostream& operator<<(ostream& O, const unordered_map<T1, T2>& v) {
  for (const auto& e : v) O << e << ' ';
  return O;
}
template <class T1, class T2>
ostream& operator<<(ostream& O, const map<T1, T2>& v) {
  for (const auto& e : v) O << e << ' ';
  return O;
}
template <class T>
istream& operator>>(istream& I, deque<T>& v) {
  for (T& e : v) I >> e;
  return I;
}
template <class T>
ostream& operator<<(ostream& O, const deque<T>& v) {
  for (const T& e : v) O << e << ' ';
  return O;
}
void _main() {
  long long n, x;
  cin >> n >> x;
  string str;
  cin >> str;
  vector<long long> pre(n + 1);
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (str[i] == '0')
      cnt++;
    else
      cnt--;
    pre[i + 1] = cnt;
  }
  if (!x && !pre[n]) {
    cout << "-1\n";
    return;
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (!pre[n]) {
      if (pre[i] == x) {
        cout << "-1\n";
        return;
      }
    } else {
      if ((x - pre[i]) % pre[n] == 0 && (x - pre[i]) / pre[n] >= 0) ans++;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _t = 1;
  cin >> _t;
  while (_t--) _main();
  return 0;
}