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 (2.7.1) (judge verdict: Wrong Answer)

MOD = 1_000_000_007
n = gets.chomp.to_i
a = gets.chomp.split.map(&:to_i)

lim = 60
bit_count = Array.new(lim, 0)
de = false

a.each do |num|
  lim.times do |i|
    bit_count[i] += num[i]
  end
end

p ['init', bit_count] if de

ans = 0

a.each_with_index do |num, index|
  lim.times do |i|
    bit_count[i] -= num[i]
  end

  p ['u bit', bit_count.reverse] if de
  new_bit_count = bit_count.dup
  lim.times do |i|
    if num[i] == 1
      new_bit_count[i] = n - 1 - index - new_bit_count[i]
    end
  end
  p ['r bit', new_bit_count.reverse] if de

  new_sum = 0
  carry = 0
  lim.times do |i|
    count = new_bit_count[i] + carry
    new_sum += (count % 2) * (1 << i)
    carry = count / 2
  end

  ans = (ans + new_sum) % MOD
  p [new_sum, ans] if de
end

puts ans