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)

N = gets.to_i
as = gets.split.map(&:to_i)
d_c = Hash.new { 0 }
MOD = 10 ** 9 + 7

def mod_pow(a, n, mod)
  ret = 1
  while n > 0
    if n[0] == 1
      ret = ret * a % mod
    end
    a = a * a % mod
    n = n >> 1
  end
  ret
end

as.each do |a|
  (0..60).each do |i|
    if a[i] == 1
      d_c[i] += 1
    end
  end
end
ans = 0
d_c.each do |d, c|
  zero_c = N - c
  n = mod_pow(2, d, MOD)
  ans = (ans + ((zero_c * c % MOD) * n % MOD)) % MOD
end
puts ans