#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, a[N];
unordered_map<int, int> ma;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  int fp = 0;
  long long d = 0, e = 0;
  for (int i = 0; i < 31; ++i) {
    ma.clear();
    long long g = 0;
    for (int j = 1; j <= n; ++j) {
      int aji = (a[j] >> i);
      ++ma[aji];
      g += ((aji & 1) ? 1 : -1) * ma[aji ^ 1];
      d += ((aji & 1) ? -0 : 1) * ma[aji ^ 1];
    }
    if (g < 0) {
      e += g;
      fp |= (1 << i);
    }
  }
  cout << d + e << " " << fp << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int N = 300005;
int n, a[N];
unordered_map<int, int> ma;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  int fp = 0;
  long long d = 0, e = 0;
  for (int i = 0; i < 31; ++i) {
    ma.clear();
    long long g = 0;
    for (int j = 1; j <= n; ++j) {
      int aji = (a[j] >> i);
      ++ma[aji];
      g += ((aji & 1) ? 1 : -1) * ma[aji ^ 1];
      d += ((aji & 1) ? -0 : 1) * ma[aji ^ 1];
    }
    if (g < 0) {
      e += g;
      fp |= (1 << i);
    }
  }
  cout << d + e << " " << fp << endl;
  return 0;
}

```