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 Scala (2.11.7) (judge verdict: Accepted)

import java.io.{BufferedReader, InputStreamReader}

object ModN {
  implicit def toModN(n: Long): ModN = {
    new ModN(n)
  }
}

class ModN(n: Long) {
  val MOD = 1000000007L

  def +(other: ModN): ModN = {
    new ModN((n + other.toLong) % MOD)
  }

  def *(other: ModN): ModN = {
    new ModN((n * other.toLong) % MOD)
  }

  val toLong: Long = n % MOD
}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val n = in.readLine().toInt
    val as = in.readLine().split("\\s").map(_.toLong)

    val cs = Array.ofDim[Int](60)

    as.foreach { a =>
      (0 until 60).foreach { shift =>
        cs(shift) = cs(shift) + ((a >> shift) & 1).toInt
      }
    }

    val ans = (0 until 60).foldLeft(new ModN(0L)) { case (acc, shift) =>
      acc + (0 until shift).foldLeft(new ModN(cs(shift)) * new ModN(n - cs(shift))) { case (acc2, _) =>
        acc2 * 2L
      }
    }

    System.out.println(ans.toLong)
  }
}