Dima got into number sequences. Now he's got sequence a1, a2, ..., an, consisting of n positive integers. Also, Dima has got a function f(x), which can be defined with the following recurrence:

  * f(0) = 0; 
  * f(2·x) = f(x); 
  * f(2·x + 1) = f(x) + 1. 



Dima wonders, how many pairs of indexes (i, j) (1 ≤ i < j ≤ n) are there, such that f(ai) = f(aj). Help him, count the number of such pairs. 

Input

The first line contains integer n (1 ≤ n ≤ 105). The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

The numbers in the lines are separated by single spaces.

Output

In a single line print the answer to the problem.

Please, don't use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3
1 2 4


Output

3


Input

3
5 3 1


Output

1

Note

In the first sample any pair (i, j) will do, so the answer is 3.

In the second sample only pair (1, 2) will do.
import java.util.Scanner;

public class Dimasequence
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int intnum;
		intnum = sc.nextInt();
		
		long[] mainnums = new long[intnum];
		for(int k =0; k<=intnum-1; k++)
			mainnums[k] = sc.nextLong();
		
		int needed = 0;
		for(int i=1; i<=intnum-1; i++)
		{
			for(int j=i; j<=intnum-1; j++)
			{
				if(mainnums[j]==mainnums[i-1])
				{
					needed+=1;
					//System.out.println("done");
				}else if(mainnums[i-1]==1)
				{
					if(mainnums[j]%2==0)
					{
						needed+=1;
						//System.out.println("don");
					}
				}else if(mainnums[j]==1)
				{
					if(mainnums[i-1]%2==0)
					{
						needed+=1;
						//System.out.println("do");
					}
				}else if((mainnums[i-1]/2)%(mainnums[j]/2)==0)
				{
					if(((mainnums[i-1]/2)/(mainnums[j]/2))/2==0)
					{
						needed+=1;
						//System.out.println("d");
					}
				}else if((mainnums[j]/2)%(mainnums[i-1]/2)==0)
				{
					if(((mainnums[j]/2)/(mainnums[i-1]/2))/2==0)
					{
						needed+=1;
						//System.out.println("d");
					}
				}
			}
			
	/*		
			
			
			
				if(mainnums[j]==mainnums[i-1])
				{
					needed+=1;
				}else if((mainnums[i-1])%2==0)
				{
					if ((mainnums[j]%2==0)||(mainnums[i-1]%mainnums[j]==0)||(mainnums[j]%mainnums[i-1]==0))
					{
						needed+=1;
						//System.out.println("done");
					}
				}else if((mainnums[i-1]%2==1))
				{
					if(mainnums[i-1]==1)
					{
						if(mainnums[j]%2==0)
							needed+=1;
					}else if{
						
					}
				}
			}*/
		}
		
		System.out.println(needed);
		
	}
}