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.*;
import java.lang.*;
import java.math.*;
public class Main extends Thread  {
    boolean[] prime;
    FastScanner sc;
    PrintWriter pw;
    long startTime = System.currentTimeMillis();
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public Main(ThreadGroup t,Runnable r,String s,long d )
    {
        super(t,r,s,d);
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
        solve();
        pw.flush();
        pw.close();
    }
    public static void main(String[] args)
    {
        new Main(null,null,"",1<<20).start();
    }
 
 
    /////////////------------------------------------//////////////
    ////////////------------------Main-Logic--------//////////////
    ///////////-------------------------------------//////////////
 
 
    public void solve() {
       int t=sc.ni();
       while(t-->0)
       {
        int n=sc.ni();
        long p=sc.ni();
        char[] arr=sc.next().toCharArray();
        HashMap<Long,Long> map=new HashMap();
        long x=0,y=0,z=-1;
        for(int i=0;i<n;i++)
        {
            if(arr[i]=='0')
            x++;
            else
            y++;
            if(map.containsKey(x-y))
            map.replace(x-y,map.get(x-y)+1L);
            else
            map.put(x-y,1L);
        }
        long res=x-y;
        long ans=0;
        if(p==0)
        ans=1;
        if(res==0)
        {
            if(map.containsKey(p))
            ans=-1;
        }
        else
        {
            
        Iterator itr=map.keySet().iterator();
        while(itr.hasNext())
        {
            long x1=(long)itr.next();
            if(((((p-x1)>0)&&(res>0))||(((p-x1)<0)&&(res<0)))||(p==x1))
            {
                long f=Math.abs(p-x1);
                long g=Math.abs(res);
                if(f%g==0)
                ans+=map.get(x1);
            }
        }
       }
       pw.println(ans);
 
}
}
    
}