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 REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i <= n; i++)
#define FORR(i, m, n) for(int i = m; i >= n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.rbegin(), v.rend());
#define ALL(v) (v).begin(),(v).end()
 
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;

const ll MOD = 1e9 + 7;

int main(){
	cin.tie(0);
    ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    ll a;
    vll one(62);
    REP(i, n) {
        cin >> a;
        ll d = 0;
        while (a) {
            one[d++] += (a & 1LL);
            a >>= 1;
        }
    }
    ll res = 0, k = 1;
    REP(i, 62) {
        res += k * one[i] * (n - one[i]);
        res %= MOD;
        k = (2 * k) % MOD;
    }
    cout << res << endl;
    return 0;
}