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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

    import java.util.*;
	import java.lang.Math;
    public class Main
	{
		//int count=0;
		/*static int xor(int x,int y)
		{
			return (x | y) & (~x | ~y);
		}*/
    	public static void main(String[] args) 
		{
    		Scanner scan = new Scanner(System.in);
			int n = scan.nextInt();
			long[] a = new long[n];
			for(int i=0;i<n;i++)
			{
				a[i]=scan.nextLong();
			}
			long mod = (long) Math.pow(10,9) +7;
			
			long ans = 0;
			long base = 1;
			for (int i = 0; i <= 60; i++)
			{
				long zeroNum = 0;
				long oneNum = 0;
				for (int j = 0; j < n; j++)
				{
					long bit = ((a[j] >> i) & 1);
 
					if (bit == 0) 
						zeroNum++;
					else 
					oneNum++;
				
			}
 
			ans += (zeroNum * oneNum % mod) * (base % mod) % mod;
			ans %= mod;
			base *= 2;
		}
 
		System.out.println(ans);
 
    	}
    }