#include <bits/stdc++.h>
using namespace std;
const int MaxN = 3e5 + 5;
long long _10[MaxN], _01[MaxN];
vector<long long> vv;
void dfs(vector<long long> v, int d) {
  if (v.empty()) return;
  if (d < 0) return;
  long long base = 1LL << d;
  vector<long long> l, r;
  long long _0 = 0, _1 = 0;
  for (int i = 0; i < v.size(); i++) {
    if ((v[i] & base) == 0) {
      _10[d] += _1;
      ++_0;
      l.push_back(v[i]);
    } else {
      _01[d] += _0;
      ++_1;
      r.push_back(v[i]);
    }
  }
  dfs(l, d - 1);
  dfs(r, d - 1);
}
void solve() {
  int n;
  cin >> n;
  vv.clear();
  long long tp;
  for (int i = 0; i < 60; i++) _01[i] = _10[i] = 0;
  for (int i = 0; i < n; i++) {
    cin >> tp;
    vv.push_back(tp);
  }
  dfs(vv, 60);
  long long cnt = 0;
  long long x = 0;
  for (int i = 60; i >= 0; i--) {
    if (_01[i] < _10[i]) {
      cnt += _01[i];
      x |= (1 << i);
    } else {
      cnt += _10[i];
    }
  }
  cout << cnt << " " << x << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 MaxN = 3e5 + 5;
long long _10[MaxN], _01[MaxN];
vector<long long> vv;
void dfs(vector<long long> v, int d) {
  if (v.empty()) return;
  if (d < 0) return;
  long long base = 1LL << d;
  vector<long long> l, r;
  long long _0 = 0, _1 = 0;
  for (int i = 0; i < v.size(); i++) {
    if ((v[i] & base) == 0) {
      _10[d] += _1;
      ++_0;
      l.push_back(v[i]);
    } else {
      _01[d] += _0;
      ++_1;
      r.push_back(v[i]);
    }
  }
  dfs(l, d - 1);
  dfs(r, d - 1);
}
void solve() {
  int n;
  cin >> n;
  vv.clear();
  long long tp;
  for (int i = 0; i < 60; i++) _01[i] = _10[i] = 0;
  for (int i = 0; i < n; i++) {
    cin >> tp;
    vv.push_back(tp);
  }
  dfs(vv, 60);
  long long cnt = 0;
  long long x = 0;
  for (int i = 60; i >= 0; i--) {
    if (_01[i] < _10[i]) {
      cnt += _01[i];
      x |= (1 << i);
    } else {
      cnt += _10[i];
    }
  }
  cout << cnt << " " << x << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  solve();
  return 0;
}

```