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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::str::FromStr;

const MOD: u64 = 1000000007;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}

fn input_vec() -> Vec<u64> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}


fn main() {
    let n: usize = input();
    let a_vec = input_vec();
    let mut a_bit_vec: Vec<Vec<char>> = Vec::new();
    for a in a_vec {
        let bit: String = format!("{:b}", a);
        let mut bit: Vec<char> = bit.chars().collect();
        bit.reverse();
        a_bit_vec.push(bit);
    }

    let mut sum = 0;
    for i in 0..60 {
        let mut zero = 0;
        let mut one = 0;
        for a_bit in &a_bit_vec {
            if a_bit.len() > i {
                if a_bit[i] == '0' {
                    zero += 1;
                } else {
                    one += 1;
                }
            } else {
                zero += 1;
            }
        }    
        // println!("{}, {}", zero, one);
        sum += 2u64.pow(i as u32) % MOD * zero % MOD * one % MOD;
        sum %= MOD;
    }
    println!("{}", sum);
}