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.Collections.Generic;
class Program
{
    static void Main(string[] args)
    {
        //modをかける場所がわからない
        int N = int.Parse(Console.ReadLine());
        long[] A = new long[N];
        string[] str = Console.ReadLine().Split();
        for (var i = 0; i < N; i++)
        {
            A[i] = long.Parse(str[i]);
        }
        
        long ans = 0;
        int mod = (int)1e9 + 7;

        // bit列60桁分をすべて走査する
        for (int k = 0; k < 60; k++)
        {
            long x0 = 0;
            long x1 = 0;
            for (int i = 0; i < N; i++)
            {
                // k桁目をビット演算で取り出す
                var x = (A[i] >> k) & 1;
                // k桁目が 0, 1 のどちらかを数えておく
                if (x == 0) x0 += 1;
                else x1 += 1;
            }

            long p = 1;
            for (int i = 0; i < k; i++)
            {
                p *= 2;
                p %= mod;
            }
            // x0 * x1 が 0, 1 の組み合わせ数
            var count = (x0 * x1) % mod;
            ans += count * p;
            ans %= mod;
        }
    }
}