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)

lines = []
while line = gets
    lines << line.chomp.split(' ').map(&:to_i)
end

N = lines[0][0]
Mod = 10 ** 9 + 7
s2 = []
zero = "000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
for i in 1..N do
  s2[i-1] = lines[1][i-1].to_s(2)
  len_lack = 60 - s2[i-1].length
  s2[i-1] = zero[1..len_lack] + s2[i-1]
end

answer = 0
for i in 1..60 do
  count1 = 0
  for j in 1..N do
    if s2[j-1][i-1] == "1" then
      count1 += 1
    end
  end
  answer += count1 * (N - count1) * (2 ** (60 - i))
end

print answer % Mod