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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int j = 0; j < t; j++) {
            int n = scanner.nextInt(), x = scanner.nextInt();
            String s = scanner.next();
            int[] v = new int[2];
            HashMap<Integer, Integer> mp = new HashMap<>();
            mp.put(0, 1);
            for (int i = 0; i < n; i++) {
                v[s.charAt(i) - '0']++;
                mp.put(v[0] - v[1], mp.getOrDefault(v[0] - v[1], 0) + (i == (n - 1) ? 0 : 1));
            }

            if (v[0] == v[1])
                System.out.println(mp.containsKey(x) ? -1 : 0);
            else {
                int am = 0;
                for (Map.Entry<Integer, Integer> i : mp.entrySet())
                    am += ((x - i.getKey()) % (v[0] - v[1]) == 0 && ((x - i.getKey()) / (v[0] - v[1]) >= 0) ? i.getValue() : 0);
                System.out.println(am);
            }
        }
    }
}