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;
typedef long long ll;
const ll INF = 1e18;
const int inf = 1e9;
#define rep(i, a, b) for (int i = a; i < b; i++)
#define per(i, b, a) for (int i = a - 1; i >= b; i--)
using Graph = vector<vector<int>>;
using pint = pair<int, int>;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dxx[8] = {1, 1, 1, 0, 0, -1, -1, -1}, dyy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};

int main() {
    int n;
    cin >> n;
    ll zero[65] = {}, one[65] = {}; //桁ごとの数を数える
    ll ans = 0;
    rep (i, 0, n) {
        ll a;
        cin >> a;
        bitset<60> a_bit (a);
        rep (j, 0, 60) {
            if (a_bit[j]) one[j]++;
            else {
                zero[j]++;
            }
        }
    }
    ll sum = 1;
    const ll mod = 1e9 + 7;
    rep (i, 0, 60) {
        ans = (ans + one[i] * zero[i] * sum % mod) % mod;
        sum = sum * 2 % mod;
        // cout << i << " " << one[i] << "\n";
    }
    cout << ans << "\n";
}