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;
using ll=long long;
using pii  = pair<int,int>;
#define rep(i,n) for(auto i=0; i<(n); i++)
#define rep2(i,x,n) for(auto i=x; i<(n); i++)
#define MOD 1000000007
#define INF 1e+9
#define MAX_V 10

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const double PI = 3.141593;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}    //最大公約数
ll lcm(ll a,ll b) {return a * b / gcd(a, b);}   //最小公倍数
ll combination(ll n, ll r) {
  if ( r * 2 > n ) r = n - r;
  ll dividend = 1;
  ll divisor  = 1;
  for ( ll i = 1; i <= r; ++i ) {
    dividend *= (n-i+1);
    divisor  *= i;
  }
  return dividend / divisor;
}

ll n;
ll ans = 0;

int main() {
    cin >> n;
    ll a[n+10];
    rep(i,n)cin >> a[i];
    ll p = 1;
    rep(i, 61)
    {
        ll cnt = 0;
        rep(j,n){
            if((a[j]>>i)&1)
                cnt++;
        }

        ans += ((n - cnt) *cnt *p)%MOD;
        p *= 2;
    }
    cout << ans << endl;

    return 0;
}