#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  if (!v.empty()) {
    copy(v.begin(), v.end(), ostream_iterator<T>(os, " "));
  }
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  int x = 0;
  long long inv = 0;
  const int BITS = 31;
  for (int i = BITS - 1; i >= 0; i--) {
    unordered_map<int, int> z, o;
    long long zo = 0, oz = 0;
    for (int j = 0; j < n; j++) {
      int prev_mask = (((1LL << (BITS - i - 1)) - 1) << (i + 1)) & a[j];
      if ((a[j] & (1LL << i))) {
        zo += z[prev_mask];
        o[prev_mask]++;
      } else {
        oz += o[prev_mask];
        z[prev_mask]++;
      }
    }
    inv += min(zo, oz);
    if (zo < oz) {
      x ^= (1 << i);
    }
  }
  cout << inv << ' ' << x << '\n';
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).

An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.

You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.

Input

First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.

Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.

Examples

Input


4
0 1 3 2


Output


1 0


Input


9
10 7 9 10 7 5 5 3 5


Output


4 14


Input


3
8 10 3


Output


0 8

Note

In the first sample it is optimal to leave the array as it is by choosing x = 0.

In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:

  * i = 2, j = 3; 
  * i = 2, j = 4; 
  * i = 3, j = 4; 
  * i = 8, j = 9. 



In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  if (!v.empty()) {
    copy(v.begin(), v.end(), ostream_iterator<T>(os, " "));
  }
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  cin >> a;
  int x = 0;
  long long inv = 0;
  const int BITS = 31;
  for (int i = BITS - 1; i >= 0; i--) {
    unordered_map<int, int> z, o;
    long long zo = 0, oz = 0;
    for (int j = 0; j < n; j++) {
      int prev_mask = (((1LL << (BITS - i - 1)) - 1) << (i + 1)) & a[j];
      if ((a[j] & (1LL << i))) {
        zo += z[prev_mask];
        o[prev_mask]++;
      } else {
        oz += o[prev_mask];
        z[prev_mask]++;
      }
    }
    inv += min(zo, oz);
    if (zo < oz) {
      x ^= (1 << i);
    }
  }
  cout << inv << ' ' << x << '\n';
  return 0;
}

```