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: Wrong Answer)


import java.util.*;

public class Main{
	public static void main(String[] args){
		
		//long startTime = System.currentTimeMillis();
		
		Scanner sc=new Scanner(System.in);
		
		int n=sc.nextInt();
		int[] count0=new int[60];
		int[] count1=new int[60];
		
		
		for(int i=0;i<n;i++){
			long l=sc.nextLong();

			for(double j=59;j>=0;j--){
				if(l/(long)(Math.pow(2,j))==1)count1[(int)j]++;
				else count0[(int)j]++;
				//System.out.println("j:"+j+" l:"+l+" res: "+l/(long)(Math.pow(2,j)));
				l=l%(long)Math.pow(2,j);
			}
		}
		
		long ans=0;
		long div=(long)Math.pow(10, 9)+7;
		
		for(int i=0;i<60;i++){
			//System.out.println("i:"+i+" 0:"+count0[i]+" 1: "+count1[i]);
			//System.out.println(1);
			ans=(ans+(long) ((count0[i]*count1[i]))*(long)Math.pow(2, i))%div;
		}
		
		
		
		System.out.println(ans);
	}
}