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 mod = 1e9 + 7;
const int maxn = 3e5 + 5;
ll arr[maxn];
int digit[maxn], bit[maxn][65];
int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    int n; cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        int j = 0; ll tmp = arr[i];
        while (tmp) {
            bit[i][j] = (tmp & 1);
            tmp >>= 1;
            ++j;
        }
    }
    for (int i = n - 1; i >= 0; --i)
        for (int j = 0; j < 65; ++j)
            bit[i][j] += bit[i + 1][j];
    ll ans = 0;
    for (int i = 0; i < n - 1; ++i) {
        int last = n - i - 1;
        for (int j = 0; j < 65; ++j) {
            int cnt = bit[i + 1][j];
            if (arr[i] & 1)
                ans = (ans + (1LL << j) * (ll)(last - cnt)) % mod;
            else
                ans = (ans + (1LL << j) * (ll)(cnt)) % mod;
            arr[i] >>= 1;
        }
    }
    cout << ans << '\n';
    return 0;
}