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: Accepted)

#include <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define gcd __gcd
#define F first
#define B begin()
#define E end()
#define ln length()
#define sz size()
#define S second
#define em empty()
#define mp make_pair
#define pb push_back
#define pq priority_queue
#define um unordered_map
#define us unordered_set
#define lb lower_bound
#define ub upper_bound
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define sbits(a) __builtin_popcount(a)
#define lcm(a, b) ((a * b) / gcd(a, b))
#define nani(x) cerr << #x << " is " << x << endl;
#define adde(g, u, v) g[u].pb(v), g[v].pb(u);
#define addew(g, u, v, w) g[u].pb({ v, w }), g[v].pb({ u, w });
#define fast ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define inf 10000000000
#define mod 1000000007

typedef long long ll;

const int N = 3e5 + 5;

ll A[N];

ll s[61], u[61];

int main()
{
    fast;

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    int n;
    cin >> n;

    for (int i = 0; i < n; i++)
        cin >> A[i];

    for (int j = 0; j < n; j++) {
        for (ll i = 0; i < 61; i++) {
            if (A[j] & (1LL << i))
                s[i]++;
            else
                u[i]++;
        }
    }

    ll ans = 0;

    for (ll i = 0; i < 61; i++) {
        ans = (ans + (((s[i] * u[i]) % mod) * ((1LL << i) % mod)) % mod) % mod;
    }

    cout << ans;

    return 0;
}