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)

MOD = 10 ** 9 + 7

n = gets.to_i
a = gets.chomp.split(" ").map(&:to_i)
bit = a.max.to_s(2).length
a.map!{|x|
  x.to_s(2)
}
(bit-1).times{
  for i in 0 .. n-1
    if a[i].length < bit
      a[i] = "0" + a[i]
    end
  end
}

ans = 0
num0 = Array.new(bit,0)
num1 = Array.new(bit,0)
for i in 0..n-1
  for j in 0..bit-1
    if a[i][j] == "1"
      num1[j] += 1
    else
      num0[j] += 1
    end
  end
end

for i in 0..bit-1
  ans += ((num0[i] * num1[i]) * (2 ** (bit-1-i))) % MOD
end
puts ans % MOD