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 <iostream>
#include<iomanip>
#include <map>
#include <algorithm>
#include <vector>
#include <string>
#include <utility>
#include <queue>
#define INF 1e9+7
#define rep(i,n) for(ll i=0;i<n;i++)
#define NO cout<<"NO"<<endl;
#define YES cout<< "YES"<<endl;
#define No cout<< "No"<<endl;
#define Yes cout<< "Yes"<<endl;
#define all(a) a.begin(),a.end()
#define P pair<int,int>
#define space ' '
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

//(int)'a'は97 (int)'A'は65 (int)'1'は49
//おまじない

const int maxn = 300000;

int main(){
    
    int n;
    ull mod = INF;
    ull ans=0;
    ull a[maxn];
    cin >> n;
    rep(i,n)cin >> a[i];
    
    ull b = 1;
    rep(i,61){
        ll c = b;
        b=b*2;
        ll A=0,B=0;
        rep(j,n){
            ll t = a[j]%b;
            if(c<=t&&t<b)A++;
            else B++;
        }
        
        ll add = (c*A%mod*B%mod)%mod;
        ans = (ans+add)%mod;
        if(ans<0)ans+=mod;
    }
    
    cout << ans<<endl;
    
    return 0;
}