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 (1.42.0) (judge verdict: Wrong Answer)

use proconio::{input, fastout};

const MOD: u64 = 1000000007;

fn modpow(a: u64, n: u64) -> u64 {
    let mut res: u64 = 1;
    let mut b: u64 = a;
    let mut m: u64 = n;
    loop{
        if m == 0 {
            break;
        }
        if m % 2 == 1 {
            res = (res * b) % MOD;
        }
        b = (b * b) % MOD;
        m = m >> 1;
    }
    res
}

#[fastout]
fn main() {
    input! {
        n: usize,
    }
    let mut v: Vec<usize> = vec![0; 64];
    for _ in 0..n {
        input! {
            mut a: u64,
        }
        for i in 0..64 {
            if a % 2 == 1 {
                v[i] += 1;
            }
            a = a >> 1;
        }
    }

    let mut ans: u64 = 0;

    for i in 0..64 {
        ans = (ans + (modpow(2,i) * (v[i as usize] as u64) * ((n - v[i as usize]) as u64) % MOD)) % MOD;
    }

    println!("{}",ans);

}