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 <typename T, typename V>
ostream &operator<<(ostream &out, const pair<T, V> x) {
  out << "{" << x.first << " : " << x.second << "}";
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const set<T> x) {
  for (auto &it : x) {
    out << it << " ";
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const multiset<T> x) {
  for (auto &it : x) {
    out << it << " ";
  }
  return out;
}
template <typename T, typename V>
ostream &operator<<(ostream &out, const map<T, V> x) {
  for (auto &it : x) {
    out << "[" << it.first << "]"
        << " = " << it.second << "\n";
  }
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> x) {
  for (int i = 0; i < x.size() - 1; ++i) {
    out << x[i] << " ";
  }
  out << x.back();
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<vector<T> > x) {
  for (int i = 0; i < x.size() - 1; ++i) {
    out << "[" << i << "]"
        << " = {" << x[i] << "}\n";
  }
  out << "[" << x.size() - 1 << "]"
      << " = {" << x.back() << "}\n";
  return out;
}
const long long N = 1e6 + 5;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
map<long long, long long> bal;
void solve() {
  bal.clear();
  long long n, x;
  string s;
  cin >> n >> x >> s;
  long long del = 0;
  for (auto &it : s) {
    if (it == '0')
      del++;
    else
      del--;
    bal[del]++;
  }
  if (del == 0) {
    if (bal.count(x)) {
      cout << -1;
    } else {
      if (x == 0)
        cout << 1;
      else
        cout << 0;
    }
    return;
  }
  long long ans = 0;
  if (x == 0) ans++;
  for (auto &it : bal) {
    if (del > 0) {
      if (it.first <= x && (x - it.first) % del == 0) ans += it.second;
    } else {
      if (it.first >= x && (it.first - x) % -del == 0) ans += it.second;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; ++i) {
    solve();
    cout << "\n";
  }
  return 0;
}