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)

//#define _GLIBCXX_DEBUG
#include "bits/stdc++.h"

using namespace std;

//------------------------------- Type Names -------------------------------//

using i64 = int_fast64_t;

using seika = string;
//{akari : 1D, yukari : 2D, maki : 3D} vector
template <class kizuna>
using akari = vector<kizuna>;
template <class yuzuki>
using yukari = akari<akari<yuzuki>>;
template <class tsurumaki>
using maki = akari<yukari<tsurumaki>>;
//{akane : ascending order, aoi : decending order} priority queue
template <class kotonoha>
using akane = priority_queue<kotonoha, akari<kotonoha>, greater<kotonoha>>;
template <class kotonoha>
using aoi = priority_queue<kotonoha>;

//------------------------------- Libraries ---------------------------------//

template <int p>
struct Modint
{
    int value;

    Modint() : value(0) {}
    Modint(long x) : value(x >= 0 ? x % p : (p + x % p) % p) {}

    inline Modint &operator+=(const Modint &b)
    {
        if ((this->value += b.value) >= p)
            this->value -= p;
        return (*this);
    }
    inline Modint &operator-=(const Modint &b)
    {
        if ((this->value += p - b.value) >= p)
            this->value -= p;
        return (*this);
    }
    inline Modint &operator*=(const Modint &b)
    {
        this->value = (int)((1LL * this->value * b.value) % p);
        return (*this);
    }
    inline Modint &operator/=(const Modint &b)
    {
        (*this) *= b.inverse();
        return (*this);
    }

    Modint operator+(const Modint &b) const { return Modint(*this) += b; }
    Modint operator-(const Modint &b) const { return Modint(*this) -= b; }
    Modint operator*(const Modint &b) const { return Modint(*this) *= b; }
    Modint operator/(const Modint &b) const { return Modint(*this) /= b; }

    inline Modint &operator++(int) { return (*this) += 1; }
    inline Modint &operator--(int) { return (*this) -= 1; }

    inline bool operator==(const Modint &b) const
    {
        return this->value == b.value;
    }
    inline bool operator!=(const Modint &b) const
    {
        return this->value != b.value;
    }
    inline bool operator<(const Modint &b) const
    {
        return this->value < b.value;
    }
    inline bool operator<=(const Modint &b) const
    {
        return this->value <= b.value;
    }
    inline bool operator>(const Modint &b) const
    {
        return this->value > b.value;
    }
    inline bool operator>=(const Modint &b) const
    {
        return this->value >= b.value;
    }

    // requires that "this->value and p are co-prime"
    // a_i * v + a_(i+1) * p = r_i
    // r_i = r_(i+1) * q_(i+1) + r_(i+2)
    // q == 1 (i > 1)
    // reference: https://atcoder.jp/contests/agc026/submissions/2845729
    // (line:93)
    inline Modint inverse() const
    {
        assert(this->value != 0);
        int r0 = p, r1 = this->value, a0 = 0, a1 = 1;
        while (r1)
        {
            int q = r0 / r1;
            r0 -= q * r1;
            swap(r0, r1);
            a0 -= q * a1;
            swap(a0, a1);
        }
        return Modint(a0);
    }

    friend istream &operator>>(istream &is, Modint<p> &a)
    {
        long t;
        is >> t;
        a = Modint<p>(t);
        return is;
    }
    friend ostream &operator<<(ostream &os, const Modint<p> &a)
    {
        return os << a.value;
    }
};

/*
verified @ https://atcoder.jp/contests/abc034/submissions/4316971
*/

const int MOD = 1e9 + 7;

using Int = Modint<MOD>;

//------------------------------- Dubug Functions ---------------------------//
inline void print()
{
    cout << endl;
}
template <typename First, typename... Rest>
void print(const First &first, const Rest &... rest)
{
    cout << first << ' ';
    print(rest...);
}
//------------------------------- Solver ------------------------------------//

void solve()
{
    int n;
    cin >> n;
    akari<i64> as(n);
    for (int i = 0; i < n; i++)
    {
        cin >> as[i];
    }
    Int ans = 0;
    for (int d = 0; d < 60; d++)
    {
        akari<int> cnt(2);
        Int w = 1LL << d;
        for (int i = 0; i < n; i++)
        {
            ans += w * cnt[(as[i] >> d & 1) ^ 1];
            cnt[as[i] >> d & 1]++;
        }
    }
    cout << ans << endl;
}

int main()
{
    solve();
    return 0;
}