#include <bits/stdc++.h>
using namespace std;
constexpr int INF = numeric_limits<int>::max() >> 1;
int a[300005], n;
long long f[30][2];
void dfs(int bit, vector<int> x) {
  if (bit == -1 || x.size() == 0) return;
  int g[2] = {0, 0};
  vector<vector<int>> temp(2);
  for (int i : x) {
    int u = (a[i] >> bit & 1);
    f[bit][u] += g[u ^ 1];
    g[u]++;
    temp[u].push_back(i);
  }
  dfs(bit - 1, temp[0]);
  dfs(bit - 1, temp[1]);
}
void solution() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<int> temp;
  for (int i = 1; i <= n; i++) temp.push_back(i);
  dfs(30, temp);
  int x = 0;
  long long ans = 0;
  for (int bit = 0; bit < 30; bit++) {
    if (f[bit][0] > f[bit][1]) {
      ans += f[bit][1];
      x ^= (1 << bit);
    } else
      ans += f[bit][0];
  }
  cout << ans << " " << x;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solution();
    cout << '\n';
  }
}

### Prompt

Your challenge is to write 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;
constexpr int INF = numeric_limits<int>::max() >> 1;
int a[300005], n;
long long f[30][2];
void dfs(int bit, vector<int> x) {
  if (bit == -1 || x.size() == 0) return;
  int g[2] = {0, 0};
  vector<vector<int>> temp(2);
  for (int i : x) {
    int u = (a[i] >> bit & 1);
    f[bit][u] += g[u ^ 1];
    g[u]++;
    temp[u].push_back(i);
  }
  dfs(bit - 1, temp[0]);
  dfs(bit - 1, temp[1]);
}
void solution() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<int> temp;
  for (int i = 1; i <= n; i++) temp.push_back(i);
  dfs(30, temp);
  int x = 0;
  long long ans = 0;
  for (int bit = 0; bit < 30; bit++) {
    if (f[bit][0] > f[bit][1]) {
      ans += f[bit][1];
      x ^= (1 << bit);
    } else
      ans += f[bit][0];
  }
  cout << ans << " " << x;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solution();
    cout << '\n';
  }
}

```