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;
using ll = long long;
template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return 0;
}
template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return 0;
}
void solve() {
  int N, X;
  string S;
  cin >> N >> X >> S;
  int cnt0 = 0, cnt1 = 0;
  for (int i = (int)0; i < (int)N; ++i) {
    if (S[i] == '0') {
      ++cnt0;
    } else {
      ++cnt1;
    }
  }
  if (cnt0 == cnt1) {
    cnt0 = cnt1 = 0;
    for (int i = (int)0; i < (int)N; ++i) {
      if (S[i] == '0') {
        ++cnt0;
      } else {
        ++cnt1;
      }
      if (cnt0 - cnt1 == X) {
        cout << -1 << endl;
        return;
      }
    }
    cout << 0 << endl;
  } else {
    if (cnt0 < cnt1) {
      X *= -1;
      for (int i = (int)0; i < (int)N; ++i) {
        if (S[i] == '0') {
          S[i] = '1';
        } else {
          S[i] = '0';
        }
      }
    }
    priority_queue<int> que;
    cnt0 = cnt1 = 0;
    for (int i = (int)0; i < (int)N; ++i) {
      if (S[i] == '0') {
        ++cnt0;
      } else {
        ++cnt1;
      }
      que.push(cnt0 - cnt1);
    }
    int diff = cnt0 - cnt1;
    int ans = 0;
    if (X == 0) {
      ++ans;
    }
    while (!que.empty()) {
      while (que.size() && que.top() >= X) {
        if (que.top() == X) {
          ++ans;
        }
        que.pop();
      }
      X -= ((X - que.top() + diff - 1) / diff) * diff;
    }
    cout << ans << endl;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int i = (int)0; i < (int)T; ++i) {
    solve();
  }
  return 0;
}