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;
const int mod = 1e9 + 7;
template <class T, class U>
inline void add_self(T &a, U b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
template <class T, class U>
inline void min_self(T &x, U y) {
  if (y < x) x = y;
}
template <class T, class U>
inline void max_self(T &x, U y) {
  if (y > x) x = y;
}
template <typename T>
void print(T v, bool show_index = false) {
  int w = 2;
  if (show_index) {
    for (int i = 0; i < v.size(); i++) cout << setw(w) << i << " ";
    cout << endl;
  }
  for (auto i = v.begin(); i != v.end(); i++) cout << setw(w) << *i << " ";
  cout << endl;
}
template <typename T>
void print_vv(T v) {
  if (v.size() == 0) {
    cout << "Empty" << endl;
    return;
  }
  int w = 3;
  cout << setw(w) << " ";
  for (int j = 0; j < v[0].size(); j++) cout << setw(w) << j << " ";
  cout << endl;
  for (auto i = 0; i < v.size(); i++) {
    cout << i << " {";
    for (auto j = 0; j != v[i].size(); j++) {
      cout << setw(w) << v[i][j] << ",";
    }
    cout << "}," << endl;
  }
  cout << endl;
}
template <class T, class U>
void print_m(map<T, U> m, int w = 3) {
  if (m.empty()) {
    cout << "Empty" << endl;
    return;
  }
  for (auto x : m) cout << "(" << x.first << ": " << x.second << ")," << endl;
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, n, x;
  string s;
  cin >> t;
  while (t--) {
    cin >> n >> x >> s;
    vector<int> pre(n);
    pre[0] = (s[0] == '0' ? 1 : -1);
    for (int i = int(1); i < int(n); i++)
      pre[i] = pre[i - 1] + (s[i] == '0' ? 1 : -1);
    int tot = pre[n - 1];
    bool inf_ans = false;
    int ans = 0;
    int bal = 0;
    for (int i = 0; i < n; ++i) {
      if (tot == 0) {
        if (bal == x) {
          inf_ans = true;
        }
      } else if (abs(x - bal) % abs(tot) == 0) {
        if ((x - bal) / tot >= 0) ++ans;
      }
      if (s[i] == '0')
        ++bal;
      else
        --bal;
    }
    if (inf_ans) ans = -1;
    cout << ans << endl;
  }
  return 0;
}