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: Runtime Error)

package ABC147A;
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();
		
		long div=(long)Math.pow(10, 9)+7;
		long[] A=new long[n];
		
		long ans=0;
		
		for(int i=0;i<n;i++){
			A[i]=sc.nextLong();
		}
		
		for(int i=0;i<60;i++){
			long a=0;
			int count1=0;

			for(int j=0;j<n;j++){
				if((A[j]>>i&1)==1)count1++;
				//System.out.println("j:"+j+" l:"+l+" res: "+l/(long)(Math.pow(2,j)));
			}
			a=a+count1*(n-count1);
			a=(a<<i);
			
			ans+=a;
			ans%=div;
		}

		
		System.out.println(ans);
		long endTime = System.currentTimeMillis();
		//System.out.println("処理時間：" + (endTime - startTime) + " ms");
	}
}