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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

	final static int r = 1000000000 + 7;
	
	public static int digitBit(long a) {
		if(a < 0)return -1;
		int ans = 1;
		while(a > 1){
			a >>= 1;
			++ans;
		}
		return ans;
	}

	public static long solve(long a[], long max) {
		long ans = 0, tmp = 0;
		int maxdigit = digitBit(max), n = a.length;
		long current = 1;
		for(int i = 0; i <= maxdigit; ++i){
			int cnt = 0;
			for(int j = 0; j < n; ++j){
				long target = a[j];
				tmp = current & target;
				if(tmp > 0)++cnt;
			}
			tmp = cnt * (n - cnt);
			tmp %= r;
			tmp *= current;
			tmp %= r;
			ans += tmp;
			ans %= r;
			current *= 2;
		}
		return ans;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long a[] = new long[n];
		long max = 0;
		for (int i = 0; i < n; ++i){
			a[i] = sc.nextLong();
			max = Math.max(a[i], max);
		}
		sc.close();
		long ans = solve(a, max);
		System.out.println(ans);
	}
}