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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;

public class Main {

    public static void main(String[] args) throws Exception {
        // write your code here
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine());
        while (t-->0){
            String[] input = reader.readLine().split(" ");
            long n = Long.parseLong(input[0]);
            long x = Long.parseLong(input[1]);
            long max = Integer.MIN_VALUE;
            long min = Integer.MAX_VALUE;
            HashMap<Long, ArrayList<Integer>> map = new HashMap<>();
            long a[] = new long[(int) n];
            Set<Long> se = new HashSet<>();
            String s = reader.readLine();
            long c = 0;
            for(int i = 0;i<n;i++){
                if(s.charAt(i)=='1') c--;
                else c++;
                a[i] = c;
                if(c>max) max = c;
                else if(c<min) min = c;
                if(map.containsKey(c)){
                    ArrayList<Integer> l = map.get(c);
                    l.add(i);
                    map.put(c,l);
                } else {
                    ArrayList<Integer> l = new ArrayList<>();
                    l.add(i);
                    map.put(c,l);
                }
            }
            if(c==0){
                if(map.containsKey(x)){
                    System.out.println(-1);
                } else System.out.println(0);
                continue;
            }
            int ans = 0;
            for(int i = 0;i<n;i++){
                long temp = x - a[i];
                if(temp%c==0){
                    if((x>=a[i] && c>0) || (x<=a[i]&&c<0)) ans++;
                }
            }
            if(x==0) ans++;
            System.out.println(ans);
        }
    }
}