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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.sql.Array;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }


    static class TaskD {
        public int log(long a){
            int num = 0;
            while (a > 0){
                num++;
                a>>=1;
            }
            return num;
        }
        public long binerypow(long a){
            if (a == 0) return 1;
            long v = binerypow(a/2);
            return v*v*(a%2==1?2:1);
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long ans = 0;
            int n = in.nextInt();
            int max = 1;
            long[] arr = new long[n];
            for (int i = 0; i <n ; i++) {
                arr[i] = in.nextLong();
                int v = log(arr[i]);
                if (v>max)max=v;
            }
            long [] zero = new long[max];
            long  [] one = new long[max];
            for (int i = 0; i <n; i++){
                int j = 0;
                while (j<max){
                    if ((arr[i]&1) == 1) one[j]++;
                    else zero[j]++;
                    arr[i]>>=1;
                    j++;
                }
            }

            for (int i = 0; i <max; i++) {
                if (zero[i]*one[i] > 0){
                    ans+=(((zero[i])%1000000007)*((one[i])%1000000007)*((binerypow(i))%1000000007))%1000000007;
                    ans%=(1000000000+7);
                }
            }
            System.out.println(ans);

        }



    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}