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.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner reader = new Scanner(System.in);

        int t = reader.nextInt(), n, x, res = 0, q;
        for (int i = 0; i < t; i++) {

            n = reader.nextInt();
            x = reader.nextInt();
            String input = reader.next();

            int[] r = new int[n];
            // x = kq + r, k >= 0

            if (input.charAt(0) == '0')
                r[0] = 1;
            else r[0] = -1;

            for (int j = 1; j < n; j++) {

                if (input.charAt(j) == '0')
                    r[j] = r[j - 1] + 1;
                else r[j] = r[j - 1] - 1;

            }

            q = r[n - 1];
            //System.out.println(q);

            if (q == 0) {

                boolean b = false;
                for (int j = 0; j < n; j++) {

                    if (r[j] == x) {
                        System.out.println("-1");
                        b = true;
                        break;
                    }
                }
                if (!b)
                    System.out.println("0");
            } else {

                for (int j = 0; j < n; j++) {

                    if (((x - r[j]) % q == 0) && ((x - r[j]) / q >= 0))
                        res++;
                }
                if (x == 0)
                    res++;
                System.out.println(res);
                res = 0;
            }
        }
    }
}