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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

const int mod = 1e9 + 7;

int add(int a, int b) {
  return a + b - mod * (a + b >= mod);
}

int mul(int a, int b) {
  return int(1ll * a * b % mod);
}

int power(int a, int x) {
  if (x == 0) {
    return 1;
  }
  return mul(power(mul(a, a), x >> 1), (x & 1 ? a : 1));
}

int divide(int a, int b) {
  return mul(a, power(b, mod - 2));
}

int sub(int a, int b) {
  return a - b + mod * (a < b);
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  int n;
  cin >> n;
  vector<int> cnt(62);
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    for (int j = 0; j < 62; ++j) {
      if (((a >> j) & 1) == 0) {
        ans = add(ans, mul((1 << j) % mod, cnt[j]));
      } else {
        ans = add(ans, mul((1 << j) % mod, i - cnt[j]));
      }
    }
    for (int j = 0; j < 62; ++j) {
      cnt[j] += ((a >> j) & 1);
    }
  }
  cout << ans;

  return 0;
}