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;
short board[100005] = {0};
map<int, int> val[102];
string s;
int n, x;
int main() {
  int tem, numZero, numOne, ma, mi;
  int T;
  cin >> T;
  while (T) {
    T--;
    cin >> n >> x;
    cin >> s;
    numZero = numOne = 0;
    ma = -100005;
    mi = 100005;
    for (int i = 0; i < n; ++i) {
      if (s[i] == '1')
        board[i] = 1, numOne++;
      else
        board[i] = 0, numZero++;
      tem = numZero - numOne;
      val[T][tem]++;
      if (tem > ma) ma = tem;
      if (tem < mi) mi = tem;
    }
    if (tem == 0) {
      if (x >= mi && x <= ma) {
        cout << -1 << endl;
        continue;
      } else {
        cout << 0 << endl;
        continue;
      }
    } else if (tem < 0) {
      int bef = 0;
      int cou = 0;
      if (bef == 0 && x == 0) cou++;
      if (x - mi < tem) {
        int tt = (int)(x - mi) / tem;
        bef += (tt - 1) * tem;
      }
      while (bef + ma >= x) {
        cou += val[T][x - bef];
        bef += tem;
      }
      cout << cou << endl;
    } else {
      int bef = 0;
      int cou = 0;
      if (bef == 0 && x == 0) cou++;
      if (x - ma > tem) {
        int tt = (int)(x - ma) / tem;
        bef += (tt - 1) * tem;
      }
      while (bef + mi <= x) {
        cou += val[T][x - bef];
        bef += tem;
      }
      cout << cou << endl;
    }
  }
  return 0;
}