Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.

Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers ai and aj is k-interesting. Your task is to help Vasya and determine this number.

Input

The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.

The second line contains the sequence a1, a2, ..., an (0 ≤ ai ≤ 104), which Vasya has.

Output

Print the number of pairs (i, j) so that i < j and the pair of integers ai and aj is k-interesting.

Examples

Input

4 1
0 3 2 1


Output

4


Input

6 0
200 100 100 100 200 200


Output

6

Note

In the first test there are 4 k-interesting pairs:

  * (1, 3), 
  * (1, 4), 
  * (2, 3), 
  * (2, 4). 



In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:

  * (1, 5), 
  * (1, 6), 
  * (2, 3), 
  * (2, 4), 
  * (3, 4), 
  * (5, 6). 
import java.util.Scanner;
public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		long num[] = new long [32769];
		for(int i=0;i<32769;i++)
		{
			long tmp = i;
			long count = 0;
			while(tmp!=0)
			{
				if((tmp & 1)==1)count++;
				tmp=tmp>>1;
			}
			num[i] = count;
		}
		while(sc.hasNext())
		{
			long n = sc.nextLong();
			long k = sc.nextLong();
			long map[] = new long[32769];
			long ret = 0;
			for(int i=0;i<n;i++)
			{
				map[sc.nextInt()]++; 
			}
			for(int i=0;i<10020;i++)
			{
				if(map[i]==0)continue;
				for(int j=i;j<10020;j++)
				{
					if(map[j]>0)
					{
						if(num[i^j]==k)
						{
							if(i==j)
							{
								ret=ret+map[i]*(map[i]-1)/2;
							}else{
								ret=ret+map[i]*map[j];
							}
						}
					}
				}
			}
			
			System.out.println(ret);
			
			
			
		}
		sc.close();
	}
}

	    		   						  	  	 			