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 (Clang 3.8.0) (judge verdict: Wrong Answer)

# include <cmath>
# include <iostream>
# include <vector>


using namespace std;
const long long DIV = 1e9 + 7;


int main() {
    long long n; cin >> n;
    vector<long long> a(n); for (long long i=0; i<n; i++) cin >> a[i];
    for (auto item: a) cout << item << ' '; cout << endl;

    long long max_item = -1001001001;
    for (auto item: a) max_item = max(max_item, item);
    if (max_item == 0) {
        cout << 0 << endl;
        return 0;
    }

    long long k = log2(max_item) + 1;
    vector<long long> ct(k, 0);

    for (long long t=0; t<n; t++) {
        for (long long i=k; i>0; i--) {
            if (1 & a[t]>>(i-1)) {
                ct[i-1]++;
            }
        }
    }

    long long ans = 0;
    for (long long t=k-1; t>=0; t--) {
        long long _ans = 1;

        for (int i=0; i<t; i++) {
            _ans *= 2;
            _ans %= DIV;
        }

        _ans *= ct[t] * (n - ct[t]);
        _ans %= DIV;

        ans += _ans;
    }

    ans %= DIV;
    cout << ans << endl;
    return 0;
}