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)

// Author: getharshkumar

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
const ll N = 1e6 + 6, M = 1e9 + 7;
ll a[N];
int main()
{
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n, x = 1, ans = 0;
    cin >> n;
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    for (ll i = 0; i < 61; i++)
    {
        ll c = 0;
        for (ll j = 0; j < n; j++)
            if (a[j] & (1LL << i))
                c++;
        ans = (ans + ((c * (n - c) % M) * x) % M) % M;
        x = (x * 2) % M;
    }
    cout << ans;
    return 0;
}