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 Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t > 0){
            int n = sc.nextInt(); int x = sc.nextInt();
            String s = sc.next();
            int[] a = new int[n];
            int diff = 0;
            for(int i=0;i<n;i++){
                if(s.charAt(i)=='0'){
                    diff+=1;
                }else{
                    diff-=1;
                }
                a[i] = diff;
            }

            int ans = 0;
            for(int i=0;i<n;i++){
                if(diff==0&&(x-a[i])==0){
                    ans++;
                }else if(diff==0){

                }
                else if(x == a[i]){
                    ans++;
                }
                else if((x-a[i])%diff==0 && ((x > a[i] && diff > 0) || (x < a[i] && diff < 0))){
                    ans++;
                }
            }
            if(x==0 && diff!=0){
                ans+=1;
            }
            if(diff ==0 && ans >0){
                ans = -1;
            }
            System.out.println(ans);
            t--;
        }
    }
}