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 FOR(i, a, b) for (int i = (a); i < (b); i++)

typedef long long ll;

template <int mod>
class ModInt {
 public:
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using modint = ModInt<1000000007>;

ll Xor(ll al, ll ar) {
  ll out = 0;
  int digit = 0;
  while (true) {
    if (al == 0 && ar == 0) {
      break;
    }
    int xor_ = ((al % 2) + (ar % 2)) % 2;
    out += xor_ * pow(2, digit);
    digit++;
    al /= 2;
    ar /= 2;
  }
  return out;
}

int main() {
  int n;
  cin >> n;
  vector<ll> vec(n);
  FOR(i, 0, n) { cin >> vec[i]; }

  modint result;
  FOR(i, 0, n - 1) {
    FOR(j, i + 1, n) { result += Xor(vec[i], vec[j]); }
  }
  cout << result.x << endl;
  return 0;
}