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)

#![allow(non_snake_case,unused_imports)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut buf_it = buf.split_whitespace();

    let N = buf_it.next().unwrap().parse::<usize>().unwrap();
    let M = 60;
    let A:Vec<_> = (0..N).map(|_| buf_it.next().unwrap().parse::<usize>().unwrap()).collect();

    let MOD:usize = 1_000_000_007;
    
    let mut C:Vec<usize> = (0..M).map(|_| 0).collect();
    for &a in &A{
        for i in 0..M{
            if (a>>i)&1==1{
                C[i] += 1;
            }
        }
    }

    let mut ans:usize = 0;
    
    for &a in &A{
        for i in 0..M{
            if (a>>i)&1==1{
                ans += ((1<<i)%MOD)*((N-C[i])%MOD)%MOD;
            } else {
                ans += ((1<<i)%MOD)*((C[i])%MOD)%MOD;
            }
            ans %= MOD
        }
    }
    ans *= (MOD+1)/2;
    ans %= MOD;
    println!("{}",ans);

}