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 <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>

using namespace std;

__int64 f(__int64 x)
{
    if(x==0)
        return 0;
    else if(x%2==0)
        return f(x/2);
    else
        return f((x-1)/2)+1;
}

vector<int> cal;
vector<int> sum;

int s(int x)
{
    int ans = 0;
    if(sum[x]!=0)
        return sum[x];
    else{
        sum[x] = sum[x-1]+(x-1);
        return sum[x];
    }
}

int main()
{
    int n;
    scanf("%d", &n);
    cal.resize(10000);
    sum.resize(10000);
    sum[2]=1;
    sum[3]=3;
    for(int i=4; i<=40; i++)
    {
        s(i);
    }
    _int64 ans = 0;
    for(int i=0; i<n; i++){
        __int64 a;
        scanf("%d", &a);
        int fa = f(a);
        cal[fa]++;
    }
    for(int i=0; i<1000; i++){
        if(cal[i]>=2)
        {
            ans+=s(cal[i]);
        }
    }
    printf("%I64d", ans);
    return 0;
}