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.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Scanner;
import java.util.AbstractMap.SimpleEntry;
import java.util.HashSet;
import java.util.Iterator;

class Main {

    static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        long[] A = new long[N + 1];
        for (int i = 1; i <= N; i++) {
            long num = sc.nextLong();
            A[i] = num;
        }
        int[] dp = new int[60];
        for (int k = 0; k < 60; k++) {
            long mask = 1l << k;
            for (int i = 1; i <= N; i++) {
                if ((A[i] & mask) != 0l) {
                    dp[k]++;
                }
            }
        }
        long ans = 0;
        for (int k = 0; k < 60; k++) {
            ans += (((long)dp[k] * (N - dp[k])) << k)%MOD;
            ans %= MOD;
            // System.out.println(ans + " " + Long.toBinaryString(ans));
        }
        System.out.println(ans);
    }
}