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 MAXM = 1e5 + 5;
const int MAXN = 1e6 + 5;
const int INF = 1e7 + 9;
const int MOD = 1e9 + 9;
const int MX = 1e9;
const int X[] = {1, -1, 0, 0}, Y[] = {0, 0, 1, -1};
vector<int> v;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    map<int, int> mp;
    int cnt = 0, ans = 0;
    for (int j = 0; j < s.size(); ++j) {
      if (s[j] == '0')
        ++cnt;
      else
        --cnt;
      mp[j] = cnt;
    }
    int k = mp[s.size() - 1];
    cnt = 0;
    bool ok = false;
    for (int j = 0; j < s.size(); ++j) {
      if (!k) {
        if (x - cnt == 0) ok = true;
      } else {
        if ((x - cnt) % k == 0)
          if ((x - cnt) / k >= 0) ++ans;
      }
      if (s[j] == '0')
        ++cnt;
      else
        --cnt;
    }
    if (ok)
      v.push_back(-1);
    else
      v.push_back(ans);
  }
  for (auto it : v) cout << it << "\n";
}