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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int mod = 1_000_000_007;
        long[] A = new long[N];
        ArrayList<int[]> a_bin = new ArrayList<int[]>();
        for (int i=0;i<N;i++) {
            A[i] = Long.parseLong(sc.next());
            int[] a = new int[62];
            long tmp = A[i];
            for (int j=0;j<62;j++) {
                if ((tmp&1)==1) {
                    a[j] = 1;
                }
                tmp=tmp>>1;
            }
            // int[] b = new int[62];
            // for (int j=0;j<62;j++) {
            //     b[j] = a[61-j];
            // }
            a_bin.add(a);
            // System.out.println(Arrays.toString(a));
        }
        int[][] dp = new int[N+1][62];
        for (int i=0;i<N;i++) {
            for (int j=0;j<62;j++) {
                dp[i+1][j] = dp[i][j]+a_bin.get(i)[j];
            }
        }
        long ans = 0L;
        for (int i=0;i<N;i++) {
            for (int j=0;j<62;j++) {
                if (a_bin.get(i)[j]==0) {
                    ans = (ans + ((long)(1L<<j)%mod*((long)dp[N][j]-dp[i+1][j])%mod))%mod;
                } else {
                    ans = (ans + ((long)(1L<<j)%mod*((long)(N-1-i)-(dp[N][j]-dp[i+1][j]))%mod))%mod;
                }
                // if (0<=j && j<3) {
                //     System.out.print(dp[N-1][j]-dp[i][j]+" ");
                //     System.out.println((N-1-i)-(dp[N-1][j]-dp[i][j]));
                // }
            }
        }
        System.out.println(ans);
        // System.out.println(Arrays.deepToString(dp));
    }
}