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: Wrong Answer)

use std::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}


fn main(){
    let n = read_as_vec::<usize>()[0];
    let a = read_as_vec::<usize>();
    let mut sum = a.iter().sum::<usize>();
    let mut bits = vec![0;64];

    for &x in &a{
        for i in 0..64{
            if (x & (1 << i)) > 0{
                bits[i] += 1;
            }
        }
    }

    let mut ans = 0;
    //for &x in a.iter().take(n-1){
    for i in 0..n-1{
        let x = a[i];
        //remove current
        for j in 0..64{
            if (x & (1 << j)) > 0{
                bits[j] -= 1;
            }
        }
        //println!("x = {}", x);
        //println!("{:?}", bits);
        for j in 0..64{
            if (x & (1 << j)) > 0{
                ans += (1 << j) * (n - i - 1- bits[j]);
            }else{
                ans += (1 << j) * bits[j];
            }

            ans %= 1_000_000_007
        }
        //println!("ans = {}", ans);

    }
    println!("{}", ans);
}