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;

#define rep(i, j) for (int i = 0; i < j; i++)
#define all(obj) (obj).begin(), (obj).end()
#define rall(obj) (obj).rbegin(), (obj).rend()
typedef long long int ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;

template<class T,class U> void chmax(T& a, U b) {if(a<b)a=b;}
template<class T,class U> void chmin(T& a ,U b) {if(b<a)a=b;}

int main() {

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(10);
    cout << fixed;

    int n;
    cin >> n;
    vll a(n);
    rep(i, n) cin >> a[i];

    const ll mod = 1e9 + 7;

    ll ans = 0;
    map<ll, ll> mp;
    for(int i = n - 1; i >= 0; i--) {
        rep(j, 60) {
            bool isone = (a[i] & (1LL << j)) > 0;
            if (isone)  {
                ans += (1LL << j) * (n - i - mp[j] - 1);
                ans %= mod;
                mp[j]++;
            } else {
                ans += (1LL << j) * mp[j];
                ans %= mod;
            }
        }
    }

    cout << ans << endl;
    return 0;
}