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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out){
            int t  =in.nextInt();
            for (int d = 0; d <t; d++) {
                int n = in.nextInt();
                int x = in.nextInt();
                String s = in.next();
                int[] diff = new int[n];
                int ans = 0;
                if (x==0)ans++;
                int z = 0;
                int o = 0;
                for (int i = 0; i <s.length() ; i++) {
                    if (s.charAt(i)=='0')z++;
                    else o++;
                    diff[i]=z-o;
                    if (diff[i]==x)ans++;
                }
                if (s.charAt(0)=='0')z++;
                else o++;
                int k = (z-o)-diff[0];
                if (ans > 0 && k==0) System.out.println(-1);
                else if (k==0) System.out.println(ans);
                else {
                    for (int i = 0; i <n; i++) {
                        if ((x-diff[i]) % k == 0 &&(x-diff[i])/k > 0)ans++;
                    }
                    System.out.println(ans);
                }



            }


        }




    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt(){return  Integer.parseInt(next());}
        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}