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;

typedef long long ll;
#define pa pair<int, int> 
#define pal pair<long long, long long>
#define pad pair<double, double>
#define pb push_back
#define mp make_pair
#define VIN(v) for(int64_t i=0;i<(v).size();i++){cin >> (v).at(i);}

typedef vector<bool> bvec;
typedef vector<int> ivec;
typedef vector<long long> lvec;
typedef vector<double> dvec;
typedef vector<pa> pvec;

typedef vector< vector<bool> > bmat;
typedef vector< vector<int> > imat;
typedef vector<string> svec;
typedef vector< vector<string> > smat;
const ll infll = (1LL << 60) - 1;
const int inf = (1 << 30) - 1;

ll gcd(ll x, ll y){
    ll r = x%y;
    if(r == 0) return y;
    else return gcd(y, r);
} 

ll lcm(ll x, ll y){
    return x*y/gcd(x,y);
}

ll getnum(ivec &x){
    int y=0;
    for (size_t i = 0; i < 60; i++)
    {
        y+=(1ll << i)*x[i];
    }
    return y;

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, dm=60;
    cin >> n;
    lvec a(n);
    imat s(dm);
    //ivec ss(60, 0);
    ivec ss;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < dm; i++){
        s[i].push_back((a[0] >> i) % 2);
        for (int j=1; j<n; j++){
            int d = (a[j] >> i) % 2;
            s[i].push_back(s[i][j-1] + d);
        }
    }
    for (auto &x : s){
        ss.push_back((n - x[n-1])*x[n-1]);
    }

    /*
    for (int i = 0; i < n-1; i++){
        for (int j=0; j<dm; j++){
            if((a[i] >> j)%2) ss[j] += (n-1-i) - (s[j][n-1] - s[j][i]);
            else ss[j] += (s[j][n-1] - s[j][i]);
        }
    }
    */

    ll d = 1e9 + 7;
    ll ans=0;
    for (int i = 0; i < dm; i++)
    {   
        ans += ((1ll << i)%d * ss[i]%d)%d;
    }
    
    cout << ans%d <<endl;

}