#include <bits/stdc++.h>
using namespace std;
int cnt[300010 * 31], f[300010 * 31][2];
long long dp[300010][2];
int n, i, x, idx;
void add_trie(int x) {
  int rad = 0, val;
  for (int bit = 30; bit >= 0; bit--) {
    if (x & (1 << bit))
      val = 1;
    else
      val = 0;
    if (!f[rad][val]) f[rad][val] = ++idx;
    dp[bit][val] += cnt[f[rad][val ^ 1]];
    rad = f[rad][val];
    cnt[rad]++;
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    add_trie(x);
  }
  long long sol = 0, nr = 0;
  for (int bit = 30; bit >= 0; bit--) {
    if (dp[bit][1] < dp[bit][0]) {
      nr += dp[bit][1];
      sol += (1 << bit);
    } else
      nr += dp[bit][0];
  }
  cout << nr << " " << sol;
  return 0;
}

### Prompt

Please formulate a Cpp solution 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;
int cnt[300010 * 31], f[300010 * 31][2];
long long dp[300010][2];
int n, i, x, idx;
void add_trie(int x) {
  int rad = 0, val;
  for (int bit = 30; bit >= 0; bit--) {
    if (x & (1 << bit))
      val = 1;
    else
      val = 0;
    if (!f[rad][val]) f[rad][val] = ++idx;
    dp[bit][val] += cnt[f[rad][val ^ 1]];
    rad = f[rad][val];
    cnt[rad]++;
  }
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    add_trie(x);
  }
  long long sol = 0, nr = 0;
  for (int bit = 30; bit >= 0; bit--) {
    if (dp[bit][1] < dp[bit][0]) {
      nr += dp[bit][1];
      sol += (1 << bit);
    } else
      nr += dp[bit][0];
  }
  cout << nr << " " << sol;
  return 0;
}

```