Problem:
Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll arr[500009];
ll sumXOR(ll n)
{
    ll sum1 = 0;
    for(ll i = 0; i <62; i++)
    {
        ll zc = 0, oc = 0;
        ll idsum = 0;
        for (ll j=0; j<n; j++)
        {
            if(arr[j]%2==0)
            {
                zc++;
            }
            else
            {
                oc++;
            }
            arr[j]/=2;
        }
        idsum =(oc*zc*((1ll<<i)%1000000007))%1000000007;

        sum1=(sum1+idsum)%1000000007;
    }
    return sum1;
}

int main()
{
    ll sum =0,n;
    cin>>n;
    for(ll i=0; i<n; i++)
    {
        cin>>arr[i];
    }
    sum = sumXOR(n);
    cout << sum<<endl;
    return 0;
}