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.*;
public class Main {

    public static void main(String[] args) {
	    Scanner in = new Scanner(System.in);
	    int t = Integer.parseInt(in.nextLine());
	    int n, x, k, cs;
	    int[] p;
	    String[] s;
	    for (int i = 0; i < t; i++){
            s = in.nextLine().split(" ");
            n = Integer.parseInt(s[0]);
            x = Integer.parseInt(s[1]);
            s = in.nextLine().split("");
            k = 0;
            cs = 0;
            boolean f;
            p = new int[n];
            for(int j = 0; j < n; j++) {
                if (s[j].equals("1"))
                    cs--;
                else
                    cs++;
                p[j] = cs;
            }
            if (cs == 0){
                f = false;
                for(int j = 0; j < n; j++)
                    if(x == p[j] || x == 0)
                        f = true;
                if (f)
                    System.out.println("-1");
                else
                    System.out.println("0");
            }
            else{
                if(x == 0){
                    k = 1;
                    for(int j = 0; j < n; j++)
                        if(((x - p[j]) % cs == 0 && (((x - p[j]) > 0 && cs > 0) || ((x - p[j]) < 0 && cs < 0))) || (p[j] == 0))
                            k++;
                }
                else{
                    for(int j = 0; j < n; j++){
                        if((x - p[j]) % cs == 0 && (((x - p[j]) > 0 && cs > 0) || ((x - p[j]) < 0 && cs < 0)))
                            k++;
                        else if (p[j] == x)
                            k++;
                    }
                }
                System.out.println(k);
                //(((x + p[j]) % cs == 0) && (((x + p[j]) > 0 && cs > 0) || ((x + p[j]) < 0 && cs < 0)) && (x + p[j] != 0))
            }
        }
    }
}