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: Accepted)

#include <bits/stdc++.h>
#define ll long long int
#define MOD 1000000007
#define INF 1e18

using namespace std;

ll power(ll n, ll k);
ll inv(ll n);
ll genfact(ll n, ll k);
ll fact(ll n);
ll nck(ll n, ll k);


ll power(ll n, ll k){
  ll res = 1;
  while(k > 0){
    if (k & 1) 
      res = res * n % MOD;
    n = n * n % MOD;
    k >>= 1;
  }
  return res;
}

ll genfact(ll n, ll k){

  ll res = 1;
  k--;
  while (k >= 0){
    res = (res * (n - k)) % MOD;
    k--;
  }

  return res;
}

ll inv(ll n){
  return power(n, MOD - 2);
}

ll fact(ll n){
  return genfact(n, n - 1);
}

ll nck(ll n, ll k){
  if (k == 0)
    return 1;
  if (n <= 0)
    return 0;

  return ((genfact(n, k) % MOD) * inv(fact(k))) % MOD;
}

int main(void){
  ll n;
  cin >> n;
  
  vector<ll> a(n);
  for (int i = 0; i < n; i++){
    cin >> a[i];
  }

  vector<vector<ll>> dp(n + 1, vector<ll>(60, 0));


  for (int i = n-1; i >= 0; i--){
    for (int j = 0; j < 60; j++){
      dp[i][j] = dp[i+1][j] + ((a[i] >> j) & 1);
    }
  }


  ll res = 0;
  for (int i = 0; i < n; i++){
    for (int j = 0; j < 60; j++){
      if ((a[i] >> j) & 1){
        res = (res + power(2, j) * (n - 1 - i - dp[i + 1][j]) % MOD) % MOD;
      }
      else {
        res = (res + power(2, j) * (dp[i + 1][j]) % MOD) % MOD;
      }
    }
  }
  
  cout << res << endl;

  return 0;
}