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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
#define SUM(a)  accumulate((a).begin(),(a).end(),0)
#define SORT(a) sort(ALL(a))
#define REV(a) reverse(ALL(a))
#define dump(x)  cerr << #x << " = " << (x) << endl;
typedef long long ll;
const int INTINF = 1<<29;
const ll LLINF = 1LL<<61;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int DX[8]={ 0, 1, 0,-1, 1, 1,-1,-1};
const int DY[8]={ 1, 0,-1, 0, 1,-1, 1,-1};

long long int sumXOR(ll arr[], int n) 
{ 
    long long int sum = 0; 
    for (int i = 0; i < 64; i++)  
    { 
        //  Count of zeros and ones 
        ll zc = 0, oc = 0;  
          
        // Individual sum at each bit position 
        long long int idsum = 0;  
        for (int j = 0; j < n; j++) 
        { 
            if (arr[j] % 2 == 0) 
                zc++; 
            else
                oc++; 
            arr[j] /= 2; 
        } 
          
        // calculating individual bit sum  
        idsum = ((oc%MOD) * (zc%MOD) * (1 << i))%MOD;  
  
        // final sum     
        sum += idsum;  
    } 
    return sum % MOD; 
} 
  


int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long int sum = 0; 
  ll arr[400000] = {}; 
  int n;
  cin >> n;
  ll a;
  REP(i,n){
    cin >> a;
    arr[i] = a;
  }
  sum = sumXOR(arr, n); 
  cout << sum << endl;
}