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: Accepted)

#include <algorithm>
#include <string.h>
#include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;
typedef long long int ll;
const int mod = 1e9 + 7;
const int N = 3e5 + 10;
ll x[N], p[64], sum1[N], sum0[N];
int main()
{
    p[0] = 1;
    for (int i = 1; i < 63; i++)
        p[i] = ((p[i - 1]) * 2ll) % mod;
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> x[i];
    ll ans = 0;
    for (int i = 0; i < 63; i++)
    {
        memset(sum1, 0, sizeof(sum1));
        memset(sum0, 0, sizeof(sum0));
        for (int j = n; j >= 1; j--)
        {
            sum1[j] = sum1[j + 1];
            sum0[j] = sum0[j + 1];
            if ((x[j] >> i) & 1)
                sum1[j]++;
            else
                sum0[j]++;
            if ((x[j] >> i) & 1)
            {
                ll temp = (sum0[j + 1] % mod * p[i] % mod) % mod;
                ans = (ans % mod + temp % mod) % mod;
            }
            else
            {
                ll temp = (sum1[j + 1] % mod * p[i] % mod) % mod;
                ans = (ans % mod + temp % mod) % mod;
            }
        }
    }
    cout << ans % mod << endl;
    return 0;
}