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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Eric
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD_147 solver = new TaskD_147();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD_147 {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            long mod = 1_000_000_007;
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = in.nextLong();

            long res = 0;
        /*if (n%2==0) {
            for (int i=0; i<n; i++)
            res ^=  a[i];
        }*/

            for (int i = 0; i <= 60; i++) {
                long nbBits0 = 0;
                long nbBits1 = 0;
                for (int j = 0; j < n; j++) {
                    boolean isBit1 = (a[j] & (1L << i)) != 0;
                    if (isBit1)
                        nbBits1++;
                    else
                        nbBits0++;
                }
                long nbBits = nbBits0 * nbBits1;
                long nbCarry = 1L << i;
                nbCarry %= mod;
                //nbCarry *= (nbBits) * (nbBits-1);
                nbCarry *= nbBits;
                nbCarry %= mod;
                res += nbCarry;
                res %= mod;
            }
            out.println(res);
        }

    }
}