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 B implements Runnable {
	FastReader sc;
	PrintWriter out;
	long mod = (long)1e9 + 7;
	
	
	void solve() 
	{	
		int tc = sc.ni();
		outer: while(tc-- > 0)
		{	
			int n = sc.ni();
			long x = sc.nl();
			
			String s = sc.nln();
			
			HashMap<Integer,Integer> hm = new HashMap<>();
			
			int val = 0;
			
			int ic = 0;
			if(x == 0) ic++;
			
			for(int i = 0; i < s.length(); i++)
			{
				if(s.charAt(i) == '0')
				{
					val++;
					
				}
				else val--;
				
				if(val == x)
				{
					ic++;
				}
				hm.put(val, hm.getOrDefault(val,0)+1);
			}
			
			if(val == 0)
			{
				if(ic > 0)
				{
					out.println(-1);
				}
				else out.println(0);
				continue outer;
			}
			
			long ans = 0L;
			for(int key:hm.keySet())
			{
				long k = x-key;
				if(k%val == 0 && ((k < 0 && val < 0) || (k > 0 && val > 0)))
				{
					ans += hm.get(key);
				}
			}
			ans += ic;
			out.println(ans);
		
		}
	}
	
	class Pair implements Comparable<Pair>
	{
		
		int x;
		int y;
		
		Pair(int x,int y)
		{
			this.x = x;
			this.y = y;
		}
		
		public int compareTo(Pair o) 
		{
			
			if(x != o.x)
			{
				if(x > o.x) return 1;  //increasing order
				else return -1;	
			}
			else
			{
				if(y > o.y) return 1;  //increasing order
				else return -1;	
			}
		}
	}
	
	
	public void run() 
	{
		out = new PrintWriter(System.out);
		sc = new FastReader();
		solve();
		out.flush();
	}

	public static void main(String[] args) 
	{
		new Thread(null, new B(), "Main", 1 << 28).start();
	}

	class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int ni() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nl() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nd() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nln() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
        

		int[] nia(int n) {
			int[] ar = new int[n];
			for (int i = 0; i < n; i++)
				ar[i] = ni();
			return ar;
		}
		
		long[] nla(int n) {
			long[] ar = new long[n];
			for (int i = 0; i < n; i++)
				ar[i] = nl();
			return ar;
		}

	    
		int[][] nim(int n, int m) {
			int[][] ar = new int[n][];
			for (int i = 0; i < n; i++) {
				ar[i] = nia(m);
			}
			return ar;
		}

		long[][] nlm(int n, int m) {
			long[][] ar = new long[n][];
			for (int i = 0; i < n; i++) {
				ar[i] = nla(m);
			}
			return ar;
		}
		
		String reverse(String s)
		{
			StringBuilder r = new StringBuilder(s);
			return r.reverse().toString();
		}
		
		void merge1(int arr[], int l, int m, int r)
	    {
	        int n1 = m - l + 1;
	        int n2 = r - m;
	        int L[] = new int [n1];
	        int R[] = new int [n2];
	        for (int i=0; i<n1; ++i)
	            L[i] = arr[l + i];
	        for (int j=0; j<n2; ++j)
	            R[j] = arr[m + 1+ j];
	        int i = 0, j = 0;
	        int k = l;
	        while (i < n1 && j < n2)
	        {
	            if (L[i]<=R[j])
	            {
	                arr[k] = L[i];
	                i++;
	            }
	            else
	            {
	                arr[k] = R[j];
	                j++;
	            }
	            k++;
	        }
	        while (i < n1)
	        {
	            arr[k] = L[i];
	            i++;
	            k++;
	        }
	        while (j < n2)
	        {
	            arr[k] = R[j];
	            j++;
	            k++;
	        }
	    }
	    void sort1(int arr[], int l, int r)
	    {
	        if (l < r)
	        {
	            int m = (l+r)/2;
	            sort1(arr, l, m);
	            sort1(arr , m+1, r);
	            merge1(arr, l, m, r);
	        }
	    }
    }
}