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.*;
import java.util.*;

public class Code {
    public static void main(String[] args)throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t=sc.nextInt();
        while(t-->0) {
            int n = sc.nextInt();int x=sc.nextInt();
            String str = sc.next();

            long cnt0 = 0,cnt = 0;
            ArrayList<Long> len = new ArrayList<>();

            for(int i=0;i<n;i++) {
                if(str.charAt(i)=='0') ++cnt0;
                if(str.charAt(i)=='1') ++cnt;
            }
            long total = cnt0-cnt;
            long val=0,ans=0;
            boolean flag=false;
            for(int i=0;i<n;i++) {
                if(total==0) {
                    if(val==x)
                        flag=true;
                }
                else if((x-val)%total==0) {
                    if((x-val)/total >= 0) {
                        ans++;
                    }
                }

                if(str.charAt(i)=='0')
                    val++;
                else
                    val--;
            }
            if(flag)
                out.println("-1");
            else
                out.println(ans+" ");
        }
        out.flush();
        out.close();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader fileReader) {
            br = new BufferedReader(fileReader);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}