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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
class ProgramD
{
    static void Main(string[] args)
    {
        //入力
        long n = int.Parse(Console.ReadLine());
        long[] a = new long[n];
        long sum = 0;
        string[] input = Console.ReadLine().Split(' ');
        long mod = 1000000000 + 7;

        //代入する
        for(long i = 0; i < n;i++)
            a[i] = long.Parse(input[i]);

        //条件入力
        for(long k = 0; k < 60;k++)
        {
            long zero = 0;
            long one = 0;
            for(long i = 0; i < n;i++)
            {
                if(bitter(a[i],(int)k) == 1)
                    one++;
                else
                    zero++;

            }
            //i桁目の和を出す
            sum += one * zero * modpow(2,k,mod) % mod;
            sum %= mod;
        }

        //答え出力
        Console.WriteLine(sum);
    }

    //xのk番目のビットを出す
    static long bitter(long x, int k)
    {
        if(((x >> k) & 1) == 0)
            return 0;
        return 1;
    }

    //繰り返し二乗法
    static long modpow(long a, long n,long mod)
    {
        if(n == 0)
            return 1;
        
        if(n % 2 == 0)
        {
            long temp = modpow(a,n/2,mod);
            return temp*temp % mod;
        }

        return a * modpow(a,n-1,mod) % mod;
    }
}