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)

// https://atcoder.jp/contests/abc147/tasks/abc147_d
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define REP(i, n) FOR(i, 0, n)
#define REPR(i, n) for (int i = n - 1; i >= 0; i--)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
#define DIV(a, b) ((a - 1) / b + 1)

template <int mod> struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(long long x_) {
        if ((x = x_ % mod + mod) >= mod)
            x -= mod;
    }
    ModInt &operator+=(ModInt rhs) {
        if ((x += rhs.x) >= mod)
            x -= mod;
        return *this;
    }
    ModInt &operator-=(ModInt rhs) {
        if ((x -= rhs.x) < 0)
            x += mod;
        return *this;
    }
    ModInt &operator*=(ModInt rhs) {
        x = (unsigned long long)x * rhs.x % mod;
        return *this;
    }
    ModInt &operator/=(ModInt rhs) {
        x = (unsigned long long)x * rhs.inv().x % mod;
        return *this;
    }
    ModInt operator-() const { return -x < 0 ? mod - x : -x; }
    ModInt operator+(ModInt rhs) const { return ModInt(*this) += rhs; }
    ModInt operator-(ModInt rhs) const { return ModInt(*this) -= rhs; }
    ModInt operator*(ModInt rhs) const { return ModInt(*this) *= rhs; }
    ModInt operator/(ModInt rhs) const { return ModInt(*this) /= rhs; }
    bool operator==(ModInt rhs) const { return x == rhs.x; }
    bool operator!=(ModInt rhs) const { return x != rhs.x; }
    ModInt inv() const { return pow(*this, mod - 2); }
    friend ostream &operator<<(ostream &s, ModInt<mod> a) {
        s << a.x;
        return s;
    }
    friend istream &operator>>(istream &s, ModInt<mod> &a) {
        s >> a.x;
        return s;
    }
};
using mint = ModInt<(int)1e9 + 7>;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N;
    cin >> N;
    vector<int> v(N);
    for (auto &x : v) {
        cin >> x;
    }

    mint res = 0;
    REP(d, 60) {
        int n0 = 0, n1 = 0;
        for (auto &x : v) {
            if (x >> d & 1) {
                n1++;
            } else {
                n0++;
            }
        }
        res += (1UL << d) * n0 * n1;
    }
    cout << res << endl;
    return 0;
}