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.io.PrintStream;
import java.util.HashMap;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
import java.util.Map;
import java.io.BufferedReader;
import java.io.InputStream;

public class b {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        int n = in.nextInt();
        for (int i=0; i<n; i++) {
            TaskB solver = new TaskB();
            solver.solve(i, in, out);
        }
        out.close();
    }

    static class TaskB {
        private int n,x;
        private String s;
        private int[] bal;

        public int[] calculate_balance() {
            int[] b = new int[n];
            if (s.charAt(0) == '0') b[0] = 1; else b[0] = -1;
            
            for (int i=1; i<n; i++) 
                if (s.charAt(i)=='0') b[i] = b[i-1] + 1;
                else b[i] = b[i-1] - 1;
            return b;
        }
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            n = in.nextInt();
            x = in.nextInt(); 
            s = in.next();
            
            bal = calculate_balance();
            //out.println(Arrays.toString(bal));
            if (bal[n-1] == 0) {
                //if exist bal[i] == x --> answer -1 
                //if not bali] == x --> answer = 0
                for (int i=0; i<n; i++) 
                    if (bal[i] == x) {
                        out.println("-1");
                        return; //done  
                    }
                out.println("0");
                return; //done 
            }

            int c = 0;    
            if (x == 0) c+=1;
            for (int i=0; i<n; i++) 
                if (((x - bal[i]) % bal[n-1] == 0) && ((x - bal[i]) / bal[n -1] >=0)) c++;
            out.println(c);
        }

    }    
    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public long[] readLongArray(int n) {
            long[] ar = new long[n];
            for (int i = 0; i < n; i++) {
                ar[i] = nextLong();
            }
            return ar;
        }

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                    //tok = new StringTokenizer(in.readLine(), ", \t\n\r\f"); //adds commas as delimeter
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }
}