#include <bits/stdc++.h>
using namespace std;
const int MX = 300005;
int n, a[MX], res;
long long inv;
void solve() {
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  vector<vector<int> > v;
  v.push_back({});
  for (int i = 0; i < (n); i++) v.back().push_back(a[i]);
  for (int i = 30; i >= 0; i--) {
    long long x = 0, y = 0;
    vector<vector<int> > nex;
    for (vector<int> &a : v) {
      int cn[2] = {0};
      vector<int> f, g;
      for (int j = 0; j < (a.size()); j++) {
        if (a[j] & (1 << i)) {
          y += cn[0];
          cn[1]++;
          f.push_back(a[j]);
        } else {
          x += cn[1];
          cn[0]++;
          g.push_back(a[j]);
        }
      }
      if (f.size()) nex.push_back(f);
      if (g.size()) nex.push_back(g);
    }
    if (y < x) {
      inv += y;
      res |= 1 << i;
    } else {
      inv += x;
    }
    v = nex;
  }
  cout << inv << " " << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int MX = 300005;
int n, a[MX], res;
long long inv;
void solve() {
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> a[i];
  vector<vector<int> > v;
  v.push_back({});
  for (int i = 0; i < (n); i++) v.back().push_back(a[i]);
  for (int i = 30; i >= 0; i--) {
    long long x = 0, y = 0;
    vector<vector<int> > nex;
    for (vector<int> &a : v) {
      int cn[2] = {0};
      vector<int> f, g;
      for (int j = 0; j < (a.size()); j++) {
        if (a[j] & (1 << i)) {
          y += cn[0];
          cn[1]++;
          f.push_back(a[j]);
        } else {
          x += cn[1];
          cn[0]++;
          g.push_back(a[j]);
        }
      }
      if (f.size()) nex.push_back(f);
      if (g.size()) nex.push_back(g);
    }
    if (y < x) {
      inv += y;
      res |= 1 << i;
    } else {
      inv += x;
    }
    v = nex;
  }
  cout << inv << " " << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```