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'
mod = 10 ** 9 + 7
n = gets.to_i
a = gets.split.map(&:to_i)

# p a

size = 64

a3 = Array.new(size, 0)
a.each {|x|
  sprintf( "%0#{size}d", x.to_s(2)).chars.each_with_index do |c, i|
    if c == "1"
      a3[i] += 1
    end
  end
}
# pp a2
# a2.size.times do |i|
#   a2[0].size.times do |ii|
#     a3[ii] += a2[i][ii]
#   end
# end

# p a3

a4 = a3.map{|x|x * (a.size - x)}.reverse

# p a4
# p a4.map.with_index{|x, i| x * (2 ** i) % mod }
p a4.map.with_index{|x, i| x * (2 ** i) % mod }.inject{|m, i| (i + m) % mod} % mod

# a3 = a2.transpose.reverse
# pp a3

# a4 = a3.map{|x| cnt = x.count{|x|x==1}; cnt * (x.size - cnt) }
# p a4
# a5 = a4.map.with_index{|x, i| x * (2 ** i) % mod }
# pp a5

# p a5.inject{|m, i| (i + m) % mod} % mod
# 1 237 103715602