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: Time Limit Exceeded)

n=int(input())
l=sorted(list(map(int,input().split())))
bitlist=[]
num=l[-1]
b1=[]
while num!=0:
  b1.append(num%2)
  num//=2
bitlist.append(b1)
for i in range(2,n+1):
  num=l[-i]
  b1=[]
  while num!=0 or len(b1)<len(bitlist[0]):
    b1.append(num%2)
    num//=2
  bitlist.append(b1)
d={}
for i in range(len(bitlist[0])):
  d[i]=0
  for j in range(n):
    d[i]+=bitlist[j][i]
ans=0
table=[1]
for i in range(60):
  table.append((2*table[-1])%(10**9+7))
for i in d: 
  ans=(ans+table[i]*(n-d[i])*d[i])%(10**9+7)
print(ans)