#include <bits/stdc++.h>
using namespace std;
long long int ans[30][2];
void rec(vector<long long int> &a, long long int bit) {
  vector<long long int> nxt[2];
  vector<long long int> cnt(2, 0);
  for (long long int i = 0; i < (long long int)a.size(); ++i) {
    if (a[i] & (1 << bit)) {
      cnt[1]++;
      ans[bit][0] += cnt[0];
      nxt[1].push_back(a[i] - (1 << bit));
    } else {
      cnt[0]++;
      ans[bit][1] += cnt[1];
      nxt[0].push_back(a[i]);
    }
  }
  if (bit == 0) return;
  if (!nxt[0].empty()) rec(nxt[0], bit - 1);
  if (!nxt[1].empty()) rec(nxt[1], bit - 1);
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; ++i) cin >> a[i];
  rec(a, 29);
  long long int a1 = 0, a2 = 0;
  for (long long int i = 0; i < 30; ++i) {
    a1 += min(ans[i][0], ans[i][1]);
    if (ans[i][0] < ans[i][1]) {
      a2 += (1 << i);
    }
  }
  cout << a1 << " " << a2 << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int T = 1;
  for (long long int i = 1; i <= T; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long int ans[30][2];
void rec(vector<long long int> &a, long long int bit) {
  vector<long long int> nxt[2];
  vector<long long int> cnt(2, 0);
  for (long long int i = 0; i < (long long int)a.size(); ++i) {
    if (a[i] & (1 << bit)) {
      cnt[1]++;
      ans[bit][0] += cnt[0];
      nxt[1].push_back(a[i] - (1 << bit));
    } else {
      cnt[0]++;
      ans[bit][1] += cnt[1];
      nxt[0].push_back(a[i]);
    }
  }
  if (bit == 0) return;
  if (!nxt[0].empty()) rec(nxt[0], bit - 1);
  if (!nxt[1].empty()) rec(nxt[1], bit - 1);
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; ++i) cin >> a[i];
  rec(a, 29);
  long long int a1 = 0, a2 = 0;
  for (long long int i = 0; i < 30; ++i) {
    a1 += min(ans[i][0], ans[i][1]);
    if (ans[i][0] < ans[i][1]) {
      a2 += (1 << i);
    }
  }
  cout << a1 << " " << a2 << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int T = 1;
  for (long long int i = 1; i <= T; ++i) {
    solve();
  }
  return 0;
}

```