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)

import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
a.reverse()
mod = 7 + 10**9
leng = 60
data = [[0 for i in range(leng+1)] for j in range(n)]
#pre
for i in range(n):
    k = list(bin(a[i]))
    k.reverse()
    ll = len(k)
    for j in range(1, leng+1):
        temp = data[i - 1][j]
        if j - 1 >= ll:
            data[i][j] = temp
            continue
        if k[j - 1] == "1":
            data[i][j] = temp + 1
        else:
            data[i][j] = temp
ans = 0
a.reverse()
data.reverse()
for i in range(n-1):
    k = format(a[i], 'b')
    k = list(k.zfill(leng))
    k.reverse()
    for j in range(leng):
        if k[j] == "1":
            ans += ((n - i - 1) - data[i+1][j+1]) * (2**j)
            ans %= mod
        else:
            ans += data[i+1][j+1] * (2**j)
            ans %= mod
print(ans)