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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys

# 改行またはスペース区切りの入力をすべて読み込んでイテレータを返します。
def get_all_int():
    return map(int, open(0).read().split())

def log(*args):
    print("DEBUG:", *args, file=sys.stderr)

def int_to_bits(x):
    return [ 1 if 2**i & x != 0 else 0 for i in range(60) ]

def inv(x, MOD):
    return pow(x, MOD-2, MOD)

MOD = 10**9+7

debug = False

if debug:
    n = 50000
    A = list(range(n))
else:
    n,*A = get_all_int()




# ビットごとの出現回数 2**i
bits = [0] * 60

for a in A:
    for i in range(60):
        if 2**i & a != 0:
            bits[i] += 1

# log("bits:", bits)

ans = 0

for a in A:
    a_bits = [ 1 if 2**i & a != 0 else 0 for i in range(60) ]

    for i in range(60):
        ans += 2**i * abs(bits[i] - a_bits[i]*n)

    # a_dash = list(map(lambda a: a*(n-1), a_bits))

    # log("a_dash", a_dash)

    # bits_dash = [ bits[i] - a_bits[i] for i in range(60)]
    # log("bits_dash", bits_dash)
    # ans += sum( ( 2**i * abs(bits_dash[i]-2*a_dash[i])) for i in range(60) )

    # ans += sum( ( 2**i * abs(bits[i]-2*a_dash[i])) for i in range(60) )

    # for i in range(60):
    #     ans = (ans + 2**i * abs(bits_dash[i] - a_dash[i])) % MOD
    # log(a, ans)

print(ans//2 % MOD)