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 int long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i,n) for(int i=0; i<(int)n; i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define COUT(x) cout << x << endl;
#define PB(x) push_back(x)
#define MP make_pair
#define vint vector<int>
int gcd(int a,int b){return b?gcd(b,a%b):a;}
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const int INF = 1e9;
const int MOD = 1e9+7;

int modpow(int a, int n, int mod) {
    int res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

signed main(){
    int n; cin >> n;
    vint a(n); rep(i,n) cin >> a[i];

    int res=0;
    rep(i,60){
        int zero=0,one=0;
        rep(j,n){
            if(a[j]&(1LL<<i)) one++;
            else zero++;
        }
        //cout<<one<<' '<<zero<<endl;
        int add=(one*zero*modpow(2,i,MOD))%MOD;
        res+=add;
        res%=MOD;
    }
    cout<<res<<endl;
}