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: Accepted)

use proconio::input;

fn main() {
    input! {
        n:i64,
        a:[i64; n],
    };

    let m = 1000000007;
    let mut cnt = vec![0; 60];

    let mut ans = 0;

    for v in &a {
        for i in 0..60 {
            if (v >> i) & 1 == 1 {
                cnt[i] += 1;
            }
        }
    }

    let mut x = n;
    for v in &a {
        for i in 0..60 {
            if (v >> i) & 1 == 1 {
                ans += (1 << i) % m * (x - cnt[i]) % m;
            }
            else {
                ans += (1 << i) % m * cnt[i] % m;
            }
            ans %= m;
        }
        for i in 0..60 {
            if (v >> i) & 1 == 1 {
                cnt[i] -= 1;
            }
        }
        x -= 1;
    }

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