#include <bits/stdc++.h>
using namespace std;
int a[300005];
int64_t inv[32][2];
void recur(int bit, const vector<int>& ind) {
  if (bit < 0 || (int)ind.size() <= 1) return;
  int64_t cnt[2]{};
  vector<int> nxt[2];
  for (int i : ind) {
    int b = !!(a[i] & (1 << bit));
    inv[bit][b] += cnt[b ^ 1];
    cnt[b]++;
    nxt[b].push_back(i);
  }
  recur(bit - 1, nxt[0]);
  recur(bit - 1, nxt[1]);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> all(n);
  iota(all.begin(), all.end(), 0);
  recur(30, all);
  int ans = 0;
  int64_t inverse = 0;
  for (int i = 30; i >= 0; i--) {
    int b = (inv[i][0] > inv[i][1]);
    inverse += inv[i][b];
    ans = ans << 1 | b;
  }
  cout << inverse << ' ' << ans << '\n';
  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;
int a[300005];
int64_t inv[32][2];
void recur(int bit, const vector<int>& ind) {
  if (bit < 0 || (int)ind.size() <= 1) return;
  int64_t cnt[2]{};
  vector<int> nxt[2];
  for (int i : ind) {
    int b = !!(a[i] & (1 << bit));
    inv[bit][b] += cnt[b ^ 1];
    cnt[b]++;
    nxt[b].push_back(i);
  }
  recur(bit - 1, nxt[0]);
  recur(bit - 1, nxt[1]);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> all(n);
  iota(all.begin(), all.end(), 0);
  recur(30, all);
  int ans = 0;
  int64_t inverse = 0;
  for (int i = 30; i >= 0; i--) {
    int b = (inv[i][0] > inv[i][1]);
    inverse += inv[i][b];
    ans = ans << 1 | b;
  }
  cout << inverse << ' ' << ans << '\n';
  return 0;
}

```