#include <bits/stdc++.h>
using namespace std;
using ll = long long;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  while (cin >> n) {
    vector<int> a(n);
    for (auto &e : a) cin >> e;
    int x = 0;
    ll inversions = 0;
    const int maxbit = 30;
    vector<unordered_map<int, pair<int, int>>> ma(maxbit);
    for (int i = maxbit - 1; i >= 0; --i) {
      ll total_inv01 = 0, total_inv10 = 0;
      for (auto e : a) {
        int up_full_mask = (1 << maxbit) - (1 << (i + 1));
        int up_mask = up_full_mask & e;
        auto &[zeros, ones] = ma[i][up_mask];
        if (e & (1 << i)) {
          total_inv01 += zeros;
          ++ones;
        } else {
          total_inv10 += ones;
          ++zeros;
        }
      }
      if (total_inv01 < total_inv10) {
        x |= (1 << i);
        inversions += total_inv01;
      } else {
        inversions += total_inv10;
      }
    }
    cout << inversions << " " << x << "\n";
  }
}

### Prompt

Create a solution in Cpp for 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;
using ll = long long;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  while (cin >> n) {
    vector<int> a(n);
    for (auto &e : a) cin >> e;
    int x = 0;
    ll inversions = 0;
    const int maxbit = 30;
    vector<unordered_map<int, pair<int, int>>> ma(maxbit);
    for (int i = maxbit - 1; i >= 0; --i) {
      ll total_inv01 = 0, total_inv10 = 0;
      for (auto e : a) {
        int up_full_mask = (1 << maxbit) - (1 << (i + 1));
        int up_mask = up_full_mask & e;
        auto &[zeros, ones] = ma[i][up_mask];
        if (e & (1 << i)) {
          total_inv01 += zeros;
          ++ones;
        } else {
          total_inv10 += ones;
          ++zeros;
        }
      }
      if (total_inv01 < total_inv10) {
        x |= (1 << i);
        inversions += total_inv01;
      } else {
        inversions += total_inv10;
      }
    }
    cout << inversions << " " << x << "\n";
  }
}

```