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: Time Limit Exceeded)

#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)

a_chs = []
a_n.each_with_index do |a, i|
    a_ch = a.to_s(2).chars.reverse!
    a_chs << a_ch
    a_ch.each_with_index do |ch, dig|
        if ch == "1"
            memo[dig] += 1
        end
    end  
end

#pp a_chs
#pp memo

count = Array.new(digit_max, 0)
a_chs.each do |a_ch|
    digit_max.times do |dig|
        ch = a_ch[dig] || '0'
        if ch == '1'
            #pp (n - memo[dig])
            count[dig] += (n - memo[dig])
        else
            #pp (memo[dig])
            count[dig] += (memo[dig])
        end
    end
end

MOD = 10 ** 9 +7
sum = 0
count.each_with_index do |cnt, digit|
    sum += ((2 ** (digit)) * cnt) / 2
    sum = sum % MOD
end

puts sum