#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n;
int a[N], x;
long long ans, res[2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int j = 29; j >= 0; j--) {
    unordered_map<int, int> cnt[2];
    cnt[0].reserve(2 * n);
    cnt[1].reserve(2 * n);
    res[0] = res[1] = 0;
    for (int i = 1; i <= n; i++) {
      int val = !!((1 << j) & a[i]);
      int num = a[i] >> (j + 1);
      res[val] += cnt[!val][num];
      cnt[val][num]++;
    }
    x *= 2;
    x += (res[0] > res[1]);
    ans += min(res[0], res[1]);
  }
  cout << ans << " " << x << endl;
  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 N = 300005;
int n;
int a[N], x;
long long ans, res[2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int j = 29; j >= 0; j--) {
    unordered_map<int, int> cnt[2];
    cnt[0].reserve(2 * n);
    cnt[1].reserve(2 * n);
    res[0] = res[1] = 0;
    for (int i = 1; i <= n; i++) {
      int val = !!((1 << j) & a[i]);
      int num = a[i] >> (j + 1);
      res[val] += cnt[!val][num];
      cnt[val][num]++;
    }
    x *= 2;
    x += (res[0] > res[1]);
    ans += min(res[0], res[1]);
  }
  cout << ans << " " << x << endl;
  return 0;
}

```