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.Scanner;
class BinaryConversion {
	public static long[] decimalToBinary(long num) {
		long[] a=new long[60];
		int i=a.length-1;
		while(num!=0) {
			a[i]=num%2;
			num/=2;
			i--;
		}
		return a;
	}
	public static long binaryToDecimal(long[] a) {
		long num=0;
		for (int i = 0; i < a.length; i++) {
			num+=(a[i]*Math.pow(2, a.length-1-i));
		}
		return num;
	}
}

 class XOR {
	public static long[] xor(long num1,long num2) {
		long[] a=BinaryConversion.decimalToBinary(num1);
		long[] b=BinaryConversion.decimalToBinary(num2);
		for (int i = 0; i < a.length; i++) {
			a[i]=(a[i]+b[i])%2;
		}
		return a;
	}
}


public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        long[] num=new long[n];
        for(int i=0;i<n;i++) {
        	num[i]=sc.nextLong();
        }
        long sum=0;
        for (int i = 0; i < num.length-1; i++) {
			for (int j = i+1; j < num.length; j++) {
				long[] temp=XOR.xor(num[i], num[j]);
				sum+=BinaryConversion.binaryToDecimal(temp);
			}
		}
        sum%=(Math.pow(10, 9)+7);
        System.out.println(sum);
	}
}