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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

N = int(input())
A = list(map(int, input().split())) 

def sumXOR( arr,  n): 
    sum = 0
    for i in range(0, 64): 
  
        #  Count of zeros and ones 
        zc = 0
        oc = 0
           
        # Individual sum at each bit position 
        idsum = 0
        for j in range(0, n): 
            if (arr[j] % 2 == 0): 
                zc = (zc + 1)
                  
            else: 
                oc = (oc + 1)
            arr[j] = int(arr[j] / 2) 
          
           
        # calculating individual bit sum  
        idsum = (oc * zc * ((1 << i))) % 1000000007
   
        # final sum     
        sum = (sum + idsum)
      
    return sum
  
  
  
  
print(sumXOR(A, N)  % 1000000007)