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 Ruby (2.3.3) (judge verdict: Wrong Answer)

#require 'pp'

n = gets.chomp!.to_i
a_n = gets.chomp!.split(" ").map!(&:to_i)
    
digit_max = a_n.max.to_s(2).length
memo = [
    Array.new(digit_max, 0),
    Array.new(digit_max, n),
]

a_n.each_with_index do |a, i|
    tmp = a
    dig = 0
    while tmp != 0
        if tmp & 1 == 1
            memo[0][dig] += 1
            memo[1][dig] -= 1
        end
        tmp >>= 1
        dig += 1
    end
end

=begin
MOD = (10 ** 9) + 7
sum = 0
count = Array.new(digit_max, 0)
digit_max.times do |dig|
    s = (2 ** dig) * (memo[0][dig] * memo[1][dig])
    sum = (sum + s) % MOD
end
    
puts sum
=end