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.
#include <bits/stdc++.h>
                     
#define pb push_back
#define mp make_pair
#define ins insert
#define file(s) freopen(s".in", "r" , stdin); freopen(s".out" , "w" , stdout);
#define IOS ios_base::sync_with_stdio(false) , cin.tie(0)
#define pii pair<int , int>
#define fi first
#define se second

using namespace std;

const int N =(int)1e5 + 1;
int a[N] , n;
map<int , int>m;

int f(int x){
	if(x == 0)
	return 0;
	else{
		if(x % 2 == 0)
		return f(x / 2);
		else
		return f((x - 1) / 2) + 1;
	}	       	
}

int sum(int x){
	return (x * x + x) / 2;
}

main () {
  	    
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}
	for(int i = 1; i <= n; i++){
		m[f(a[i])]++;
	}
  	cout << sum(m[0] - 1)+ sum(m[1] - 1) + sum(m[2] - 1);
  	return 0;
}                