#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
const int N = 3e5 + 10;
int n, tot;
int trie[31 * N][2], sz[31 * N];
long long dp[31][2];
void insert(int x) {
  int rt = 0;
  for (int i = 30; i >= 0; --i) {
    int p = (x >> i) & 1;
    dp[i][p] += sz[trie[rt][p ^ 1]];
    if (trie[rt][p] == 0) trie[rt][p] = ++tot;
    rt = trie[rt][p];
    ++sz[rt];
  }
}
int main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    insert(x);
  }
  long long sum = 0;
  int res = 0;
  for (int i = 30; i >= 0; --i) {
    if (dp[i][1] < dp[i][0]) {
      res |= 1 << i;
      sum += dp[i][1];
    } else
      sum += dp[i][0];
  }
  cout << sum << " " << res << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
const int N = 3e5 + 10;
int n, tot;
int trie[31 * N][2], sz[31 * N];
long long dp[31][2];
void insert(int x) {
  int rt = 0;
  for (int i = 30; i >= 0; --i) {
    int p = (x >> i) & 1;
    dp[i][p] += sz[trie[rt][p ^ 1]];
    if (trie[rt][p] == 0) trie[rt][p] = ++tot;
    rt = trie[rt][p];
    ++sz[rt];
  }
}
int main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    insert(x);
  }
  long long sum = 0;
  int res = 0;
  for (int i = 30; i >= 0; --i) {
    if (dp[i][1] < dp[i][0]) {
      res |= 1 << i;
      sum += dp[i][1];
    } else
      sum += dp[i][0];
  }
  cout << sum << " " << res << '\n';
  return 0;
}

```