#include <bits/stdc++.h>
using namespace std;
const int64_t INF = 2e9;
int64_t v1[40] = {0}, v2[40] = {0};
void dfs(vector<int64_t>& a, int64_t k) {
  if (a.size() == 0 || k == -1) return;
  vector<int64_t> one, zero;
  int64_t o = 0, z = 0;
  for (auto first : a) {
    int64_t v = first & (1 << k);
    if (v)
      o++, v2[k] += z, one.push_back(first);
    else
      z++, v1[k] += o, zero.push_back(first);
  }
  dfs(zero, k - 1);
  dfs(one, k - 1);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  int64_t n;
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  for (int64_t i = 0; i < 40; i++) v1[i] = v2[i] = 0;
  dfs(a, 30);
  int64_t first = 0, ans = 0;
  for (int64_t i = 0; i < 31; i++) {
    if (v2[i] < v1[i])
      first ^= (1 << i), ans += v2[i];
    else
      ans += v1[i];
  }
  cout << ans << " " << first << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int64_t INF = 2e9;
int64_t v1[40] = {0}, v2[40] = {0};
void dfs(vector<int64_t>& a, int64_t k) {
  if (a.size() == 0 || k == -1) return;
  vector<int64_t> one, zero;
  int64_t o = 0, z = 0;
  for (auto first : a) {
    int64_t v = first & (1 << k);
    if (v)
      o++, v2[k] += z, one.push_back(first);
    else
      z++, v1[k] += o, zero.push_back(first);
  }
  dfs(zero, k - 1);
  dfs(one, k - 1);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  int64_t n;
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  for (int64_t i = 0; i < 40; i++) v1[i] = v2[i] = 0;
  dfs(a, 30);
  int64_t first = 0, ans = 0;
  for (int64_t i = 0; i < 31; i++) {
    if (v2[i] < v1[i])
      first ^= (1 << i), ans += v2[i];
    else
      ans += v1[i];
  }
  cout << ans << " " << first << endl;
  return 0;
}

```