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)

#![allow(unused_imports)]

use std::cmp;
use std::cmp::{max, min, Ordering};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::io::stdin;
use std::ops::*;
use std::process::exit;
use std::str::FromStr;

pub const MOD: usize = 1000000007;

fn main() {
    let _n: usize = input();
    let a: Vec<usize> = input_vec();

    let mut ans = 0usize;
    for at in 0..61 {
        let mut zero = 0usize;
        let mut one = 0usize;
        for m in a.iter() {
            if *m & (1 << at) != 0 {
                one += 1;
            } else {
                zero += 1;
            }
        }
        ans += zero * one % MOD * (1 << at) % MOD;
        ans %= MOD;
    }

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

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

pub fn input_vec<T: FromStr>() -> Vec<T> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|x| x.parse().ok().unwrap())
        .collect::<Vec<_>>()
}

pub fn input_chars() -> Vec<char> {
    let s: String = input();
    s.chars().collect()
}

pub fn input_lines<T: FromStr>(l: usize) -> Vec<T> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input());
    }
    res
}

pub fn input_lines_two<T: FromStr, U: FromStr>(l: usize) -> Vec<(T, U)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_two());
    }
    res
}

pub fn input_lines_three<T: FromStr, U: FromStr, V: FromStr>(l: usize) -> Vec<(T, U, V)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_three());
    }
    res
}

pub fn input_two<T: FromStr, U: FromStr>() -> (T, U) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
    )
}

pub fn input_three<T: FromStr, U: FromStr, V: FromStr>() -> (T, U, V) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
        s[2].parse::<V>().ok().unwrap(),
    )
}

pub fn input_four<T: FromStr, U: FromStr, V: FromStr, W: FromStr>() -> (T, U, V, W) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse().ok().unwrap(),
        s[1].parse().ok().unwrap(),
        s[2].parse().ok().unwrap(),
        s[3].parse().ok().unwrap(),
    )
}

pub fn input_matrix<T: FromStr + Clone>(h: usize) -> Vec<Vec<T>> {
    let mut res = vec![Vec::new(); h];
    for i in 0..h {
        res[i] = input_vec();
    }
    res
}