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())
a = list(map(int, input().split( )))
ans = 0
mod = 10**9+7
Bit = [[0]*2 for j in range(60)]# for i in range(n+1)

for i in range(1,n):##1から
    for j in range(60):
        if not ((a[i-1]>>j)&1):#a[i-1]のjビット目0
            Bit[j][0] += 1
        else:#ｊビット目1
            Bit[j][1] += 1
        if ((a[i]>>j)&1):#aiのjビット目1
            ans += Bit[j][0]<<j
            ans %=mod
        else:
            ans += Bit[j][1]<<j
            ans %= mod
    #print(Bit[:6])
"""
for i in range(1,n):##1から
    for j in range(60):

    #ans += a[i]
    #ans %=mod
    #自身の分
        #print(ans)
"""
print(ans)