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 Lua (5.3.2) (judge verdict: Time Limit Exceeded)

local mod = 1000000007
local function badd(x, y) return (x + y) % mod end
local function bmul(x, y) return (x * y) % mod end

local n = io.read("*n")
local box = {0}
local mul = {1}
for i = 2, 61 do
  box[i] = 0
  mul[i] = bmul(mul[i - 1], 2)
end
local ret = 0
for i = 1, n do
  local a = io.read("*n")
  for j = 1, 61 do
    local p = a % 2
    if p == 1 then
      ret = badd(ret, bmul(mul[j], i - 1 - box[j]))
    else
      ret = badd(ret, bmul(mul[j], box[j]))
    end
    box[j] = box[j] + p
    a = a // 2
  end
end
print(ret)