#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int mod = 51061;
const int maxn = 6e5 + 10;
const int inf = 0x3f3f3f3f;
int cnt = 1, ch[maxn << 4][2], val[maxn << 4];
int root = 1;
ll ans[30][2];
void ins(int k, int x, int dep) {
  if (dep < 0) return;
  int now = x & (1 << dep);
  now = !!now;
  if (!ch[k][now]) ch[k][now] = ++cnt;
  if (now) {
    ans[dep][1] += val[ch[k][0]];
  } else {
    ans[dep][0] += val[ch[k][1]];
  }
  val[ch[k][now]]++;
  ins(ch[k][now], x, dep - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = (1); i <= (int)(n); i++) {
    int x;
    cin >> x;
    ins(1, x, 30);
  }
  ll sum = 0, v = 0;
  for (int i = (0); i <= (int)(30); i++) {
    if (ans[i][0] <= ans[i][1]) {
      sum += ans[i][0];
    } else {
      sum += ans[i][1];
      v += (1 << i);
    }
  }
  cout << sum << ' ' << v << '\n';
  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;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int mod = 51061;
const int maxn = 6e5 + 10;
const int inf = 0x3f3f3f3f;
int cnt = 1, ch[maxn << 4][2], val[maxn << 4];
int root = 1;
ll ans[30][2];
void ins(int k, int x, int dep) {
  if (dep < 0) return;
  int now = x & (1 << dep);
  now = !!now;
  if (!ch[k][now]) ch[k][now] = ++cnt;
  if (now) {
    ans[dep][1] += val[ch[k][0]];
  } else {
    ans[dep][0] += val[ch[k][1]];
  }
  val[ch[k][now]]++;
  ins(ch[k][now], x, dep - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = (1); i <= (int)(n); i++) {
    int x;
    cin >> x;
    ins(1, x, 30);
  }
  ll sum = 0, v = 0;
  for (int i = (0); i <= (int)(30); i++) {
    if (ans[i][0] <= ans[i][1]) {
      sum += ans[i][0];
    } else {
      sum += ans[i][1];
      v += (1 << i);
    }
  }
  cout << sum << ' ' << v << '\n';
  return 0;
}

```