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 Ex1 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) {
            int length = in.nextInt();
            int balance = in.nextInt();
            String s = in.next();
            int val = 0;
            int min = 0;
            int max = 0;
            int counter = 0;
            for (int i = 0; i < length; i++) {
                if ((s.charAt(i) == '0')) {
                    val++;
                } else {
                    val--;
                }
                min = Math.min(val, min);
                max = Math.max(val, max);
            }
            if (val == 0 && balance >= min && balance <= max) {
                System.out.println("-1");
                continue;
            } else if (val == 0) {
                System.out.println("0");
                continue;
            }
            int totalVal = val;
            val = 0;
            if (balance == 0) {
                counter++;
            }
            for (int i = 0; i < length; i++) {
                if ((s.charAt(i) == '0')) {
                    val++;
                } else {
                    val--;
                }
                if ((balance-val) % totalVal == 0 && (balance-val)/totalVal >= 0) {
                    counter++;
                }
            }
            System.out.println(counter);

        }
    }
}