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.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int M = 1_000_000_007;
    int N = scanner.nextInt();
    long[] a = new long[N];
    long[] numOnes = new long[60];
    for (int i = 0; i < N; i++) {
      long aa = scanner.nextLong();
      a[i] = aa;
      int j = 0;
      while (aa > 0) {
        if (aa % 2 != 0) numOnes[j]++;
        j++;
        aa /= 2;
      }
    }

    long[] b = new long[60];
    for (int i = 0; i < 60; i++) {
      b[i] = modpow(2, i, M);
    }

    long res = 0;
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < 60; j++) {
        if (a[i] % 2 == 0) {
          res += numOnes[j] * b[j] % M;
        } else {
          res += (N - numOnes[j]) * b[j] % M;
        }
        if (res >= M) res -= M;
        a[i] /= 2;
      }
    }
    res *= inverse(2, M);
    res %= M;
    System.out.println(res);
  }

  private static long inverse(long a, long mod) {
    return modpow(a, mod - 2, mod);
  }

  private static long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
      if (n % 2 == 1) res = (res * a) % mod;
      a = (a * a) % mod;
      n >>= 1;
    }
    return res % mod;
  }
}