#include <bits/stdc++.h>
using namespace std;
long long int inv[32][2];
void rec(vector<int>& a, int b) {
  if (b < 0 || a.size() == 0) return;
  int c[2] = {0};
  vector<int> v[2];
  for (int x : a) {
    int val;
    if ((x & (1 << b)) > 0)
      val = 1;
    else
      val = 0;
    if (val == 0)
      inv[b][0] += c[1];
    else
      inv[b][1] += c[0];
    c[val]++;
    v[val].push_back(x);
  }
  rec(v[0], b - 1);
  rec(v[1], b - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  rec(a, 30);
  long long int x = 0, ans = 0;
  for (int i = 0; i <= 30; i++) {
    ans += min(inv[i][0], inv[i][1]);
    if (inv[i][0] > inv[i][1]) x += (1 << i);
  }
  cout << ans << " " << x << "\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;
long long int inv[32][2];
void rec(vector<int>& a, int b) {
  if (b < 0 || a.size() == 0) return;
  int c[2] = {0};
  vector<int> v[2];
  for (int x : a) {
    int val;
    if ((x & (1 << b)) > 0)
      val = 1;
    else
      val = 0;
    if (val == 0)
      inv[b][0] += c[1];
    else
      inv[b][1] += c[0];
    c[val]++;
    v[val].push_back(x);
  }
  rec(v[0], b - 1);
  rec(v[1], b - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  rec(a, 30);
  long long int x = 0, ans = 0;
  for (int i = 0; i <= 30; i++) {
    ans += min(inv[i][0], inv[i][1]);
    if (inv[i][0] > inv[i][1]) x += (1 << i);
  }
  cout << ans << " " << x << "\n";
}

```