#include <bits/stdc++.h>
using namespace std;
int n;
int a[300010];
map<int, vector<int> > g[31];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int x = 0;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) g[30][0].push_back(a[i]);
  for (int i = 29; i >= 0; --i) {
    long long inv0 = 0, inv1 = 0;
    for (map<int, vector<int> >::iterator ww = g[i + 1].begin();
         ww != g[i + 1].end(); ++ww) {
      int calc0 = 0, calc1 = 0;
      for (vector<int>::iterator it = ww->second.begin();
           it != ww->second.end(); ++it) {
        if (*it & (1 << i)) {
          ++calc1;
          inv1 += calc0;
        } else {
          ++calc0;
          inv0 += calc1;
        }
      }
    }
    if (inv0 <= inv1)
      ans += inv0, x |= (0 << i);
    else
      ans += inv1, x |= (1 << i);
    for (map<int, vector<int> >::iterator ww = g[i + 1].begin();
         ww != g[i + 1].end(); ++ww) {
      int j0 = ww->first, j1 = ww->first | (1 << i);
      for (vector<int>::iterator it = ww->second.begin();
           it != ww->second.end(); ++it) {
        if (*it & (1 << i))
          g[i][j1].push_back(*it);
        else
          g[i][j0].push_back(*it);
      }
    }
  }
  cout << ans << " " << x << 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;
int n;
int a[300010];
map<int, vector<int> > g[31];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int x = 0;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) g[30][0].push_back(a[i]);
  for (int i = 29; i >= 0; --i) {
    long long inv0 = 0, inv1 = 0;
    for (map<int, vector<int> >::iterator ww = g[i + 1].begin();
         ww != g[i + 1].end(); ++ww) {
      int calc0 = 0, calc1 = 0;
      for (vector<int>::iterator it = ww->second.begin();
           it != ww->second.end(); ++it) {
        if (*it & (1 << i)) {
          ++calc1;
          inv1 += calc0;
        } else {
          ++calc0;
          inv0 += calc1;
        }
      }
    }
    if (inv0 <= inv1)
      ans += inv0, x |= (0 << i);
    else
      ans += inv1, x |= (1 << i);
    for (map<int, vector<int> >::iterator ww = g[i + 1].begin();
         ww != g[i + 1].end(); ++ww) {
      int j0 = ww->first, j1 = ww->first | (1 << i);
      for (vector<int>::iterator it = ww->second.begin();
           it != ww->second.end(); ++it) {
        if (*it & (1 << i))
          g[i][j1].push_back(*it);
        else
          g[i][j0].push_back(*it);
      }
    }
  }
  cout << ans << " " << x << endl;
  return 0;
}

```