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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        Long[] numbers = new Long[num];

        for(int i = 0; i < num; i++)
          numbers[i] = Long.parseLong(sc.next());

        long total = 0;
        for(int i = 0; i < num-1; i++){
          for(int j = i+1; j < num; j++){
            //System.out.println(i + " " + j + " " + Integer.parseInt(xorbinary(numbers[i], numbers[j]), 2));
            total += numbers[i]^numbers[j];
          }
        }

        System.out.println(total%1000000007);
    }

    static String xorbinary(long numA, long numB){

      String aBinary = Long.toBinaryString(numA);
      String bBinary = Long.toBinaryString(numB);

      StringBuffer sb;
      if(!aBinary.substring(aBinary.length()-1,aBinary.length()).equals(bBinary.substring(bBinary.length()-1,bBinary.length()))){
        sb = new StringBuffer("1");
      }else{
        sb = new StringBuffer("0");
      }

      if(aBinary.length() == bBinary.length()){ //文字数が同じ
        for(int i = aBinary.length()-2; i >= 0; i--){
          if(!aBinary.substring(i, i+1).equals(bBinary.substring(i, i+1)))
            sb.insert(0, "1");
          else
            sb.insert(0, "0");
        }
      }else if(aBinary.length() < bBinary.length()){ //aの方が文字数が少ない時
        for(int i = aBinary.length()-2; i >= 0; i--){
          if(!aBinary.substring(i, i+1).equals(bBinary.substring(i, i+1)))
            sb.insert(0, "1");
          else
            sb.insert(0, "0");
        }

        for(int i = bBinary.length()-aBinary.length()-1; i >= 0; i--){
          if(bBinary.substring(i, i+1).equals("1"))
            sb.insert(0, "1");
          else
            sb.insert(0, "0");
        }
      }else{ //bの方が文字数が少ない時
        for(int i = bBinary.length()-2; i >= 0; i--){
          if(aBinary.substring(i, i+1).equals(bBinary.substring(i, i+1)))
            sb.insert(0, "1");
          else
            sb.insert(0, "0");
        }

        for(int i = aBinary.length()-bBinary.length()-1; i >= 0; i--){
          if(bBinary.substring(i, i+1).equals("1"))
            sb.insert(0, "1");
          else
            sb.insert(0, "0");
        }
      }

      return sb.toString();
    }
}