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;
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
typedef long long ll;
typedef unsigned long long ull;
#define FIN ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define forr(i, a, b) for(ll i = (a); i < (ll) (b); i++)
#define forn(i, n) forr(i, 0, n)
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define DBG(x) cerr << #x << " = " << (x) << endl
#define MOD 1000000007
ll binaryNum[62];
void decToBinary(ll n)
{
    forn(i,62) binaryNum[i]=0;

    int i = 0;
    while (n > 0) {
        binaryNum[i] = n % 2;
        n = n / 2;
        i++;
    }
}
ll result (ll pot, ll num)
{
    ll loqueda=num;
    while(pot>1)
    {
        loqueda=(2*loqueda)%MOD;
        pot=pot/2;
    }
    return loqueda;
}
int main() {
    FIN;
    int n;
    ll cuantos[62];
    forn(i,62) cuantos[i]=0;
    ll x;
    cin >> n;
    forn(i,n)
    {
        cin >> x;
        decToBinary(x);
        forn(i,62) cuantos[i]+=binaryNum[i];
    }
    ll resp=0, prod, pot=1;
    forn(i,62)
    {
        prod=result(pot,((cuantos[i]*(n-cuantos[i]))%MOD));
        resp=(resp+prod)%MOD;
        pot*=ll(2);
    }
    cout << resp << endl;


    return 0;
}