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 <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>
#define vv(a, b, c, d) vector<vector<d> >(a, vector<d>(b, c))
#define vvi vector<vector<int> >
#define vvl vector<vector<ll> >
#define vll vector<ll>
typedef long long int ll;
typedef long double ld;
using namespace std;

//-------------------------------------------------------------------
#define P 1000000007
#define N_MAX 2000000
typedef long long int ll;
ll fac[N_MAX+1];
ll inv[N_MAX+1];
ll finv[N_MAX+1];

ll comb(ll n, ll k){
    return (((fac[n]*finv[n-k])%P)*finv[k])%P;
}

void init(){
    fac[0] = finv[0] = fac[1] = finv[1] = inv[1] = 1;
    for(int i = 2; i <= N_MAX; i++){
        fac[i] = (fac[i-1]*i)%P;
        inv[i] = ((-(P/i)*inv[P%i])%P+P)%P;
        finv[i] = (finv[i-1]*inv[i])%P;
    }
}
ll pp(ll a, ll b){
  return (a * b)%P;
}
//-------------------------------------------------------------------
int main(int argc, char const *argv[]) {
  ll n;std::cin >> n;
  ll ans = 0;
  init();
  std::vector<ll> a(n);
  vvl d = vv(n, 62, 0, ll);
  for(int i=0;i<n;i++) std::cin >> a[i];
  for(int i=0;i<n;i++){
    ll tmp = a[i];
    ll c = 0;
    while(tmp>0){
      if(tmp%2==1) d[i][c] = 1;
      c++,tmp/=2;
    }
    //for(int j=61;j>=0;j--) std::cout << d[i][j];
    //std::cout << '\n';
  }
  std::vector<ll> rui(63, 1);
  for(int i=1;i<62;i++) rui[i] = (rui[i-1] * 2)%P;
  for(int i=0;i<62;i++){
    ll c = 0;
    for(int j=0;j<n;j++) if(d[j][i]==1) c++;
    ans = (ans + ((rui[i]*(((n-c)*c)%P))%P))%P;
  }
  std::cout << ans << '\n';
  return 0;
}