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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using P = pair<ll,ll>;
const ll mod = 1e9+7;
#define ALL(x) (x).begin(),(x).end()
#define pow(x,y) modpow(x,y)
#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define REPS(i,n) for(ll (i)=1;(i)<=(n);(i)++)
#define RREP(i,n) for(ll (i)=(n-1);(i)>=0;(i)--)
#define RREPS(i,n) for(ll (i)=(n);(i)>0;(i)--)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());

template <class T = int>T in(){
    T x;
    cin >> x;
    return (x);
}

template<class T> inline void chmin(T& a, T b){
    if (a > b){
        a = b;
    }
}

template<class T> inline void chmax(T& a, T b){
    if (a < b){
        a = b;
    }
}

/* modとる */
inline ll modu(ll a){
    return(a % mod + mod) % mod;
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(10);
    ll N = in<ll>();
    ll answer = 0;
    vector<ll> A(N);
    REP(i,N)A[i] = in<ll>();
    REP(i,62){
        ll sum = 0;
        REP(l,N){
            if(A[l] & 1LL << i)sum++;
        }
        answer = modu(answer + modu(sum*(N-sum))*modu(1LL << i));
    }
    cout << answer << "\n";
}