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.
/***********************************************\
 |Author: WiYR
 |Created Time: 2013-8-10 20:16:25
 |File Name:Dima and Sequence.cpp
 |Description: 
\***********************************************/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>

using namespace std;
int a[30];
int main() {
//	freopen("input.txt","r",stdin); 
    int n,t;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&t);
        int k=0;
        while(t){
            if(t&1)k++;
            t>>=1;
        }
        a[k]++;
    }
    __int64 ans=0;
    for(int i=1;i<30;i++)
        ans+=a[i]*(a[i]-1)/2;
    printf("%I64d\n",ans);    
	return 0;
}
