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.io.InputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();
		long[] A = new long[N];
		for(int i=0;i<N;i++) {
			A[i] = sc.nextLong();
		}

		sc.close();


		int mod = 1000000007;

		int[][] bitCntSum = new int[N+1][63]; // iより前のbitの立ってる数

		for(int i=1;i<=N;i++) {
			long n = A[i-1];

			for(int j=0;j<63;j++) {
				bitCntSum[i][j] = (int)(bitCntSum[i-1][j] + ((n>>j)&1));
			}
		}

		long ans = 0;

		for(int j=0;j<63;j++) {
			// bitごとにみる
			for(int i=0;i<N;i++) {
				long c;
				if((A[i]>>j &0x1) == 1) {
					// 1なら、それより右の0の数を数える
					c = (N-i) - (bitCntSum[N][j] - bitCntSum[i][j]);
				}else {
					// 0なら、それより右の1の数を数える
					c = bitCntSum[N][j] - bitCntSum[i][j];
				}
				ans += ((c<<j)%mod);
				ans %= mod;
			}
		}

		ans %= mod;

		out.println(ans);

	}

	public static void main(String[] args) {
		new Main()._main(args);
	}
}