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.*;
import java.util.stream.LongStream;

public class Main implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int N = Integer.parseInt(scanner.nextLine());
        long[] nums = Arrays.stream(scanner.nextLine().split(" ")).mapToLong(Long::parseLong).toArray();
        int[][] bits = new int[N + 1][61];
        int max = 0;
        for (int i = 0; i < nums.length; i++) {
            long n = nums[i];
            for (int j = 0; j <= 60; j++) {
                if ((n & (1L << j)) > 0) {
                    bits[i + 1][j] = bits[i][j] + 1;
                    max = Math.max(max, j);
                } else {
                    bits[i + 1][j] = bits[i][j];
                }
            }

        }

        long ans = 0;
        for (int i = 0; i < nums.length; i++) {
            long n = nums[i];
            for (int j = 0; j <= max; j++) {
                int cnt = bits[N][j] - bits[i + 1][j];
                if ((n & (1L << j)) > 0) {
                    ans = (ans + (((1L << j) * (N - i - 1 - cnt)) % MOD)) % MOD;
                } else {
                    ans = (ans + (((1L << j) * cnt) % MOD)) % MOD;
                }
            }
        }

        System.out.println(ans);
    }

}