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 C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <cstdio>
typedef long long lld;
const int LOGN = 60;
const int MOD = 1e9+7;
int cnt[LOGN]{ };
lld A;
int N;

int main() {
  scanf("%d", &N);
  int ret = 0;
  for (int i = 0; i < N; ++i) {
    scanf("%lld", &A);
    for (int j = 0; j < LOGN; ++j) {
      int two = (1LL << j) % MOD;
      int val;
      if ((A >> j) & 1) {
        val = 1LL * two * (i - cnt[j]) % MOD;
        ++cnt[j];
      } else {
        val = 1LL * two * cnt[j] % MOD;
      }
      ret = (1LL * ret + val) % MOD;
    }
  }
  printf("%d\n", ret);
  return 0;
}