#include <bits/stdc++.h>
using namespace std;
struct tri {
  int first, second, z;
};
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 3e5 + 10;
const long long mod = 1e9 + 7;
const double PI = acos(0) * 2;
int n, arr[40][N], brr[40][N];
int first;
long long c0[40], c1[40], ans;
void dfs(vector<int> &v, int d) {
  if (v.size() <= 1) return;
  vector<int> v0, v1;
  for (auto i : v) {
    if (i & (1 << d)) {
      v1.push_back(i);
      c1[d] += v0.size();
    } else {
      v0.push_back(i);
      c0[d] += v1.size();
    }
  }
  if (d) {
    dfs(v0, d - 1);
    dfs(v1, d - 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  dfs(v, 30);
  for (int i = 0; i <= 30; i++) {
    if (c1[i] < c0[i]) {
      ans += c1[i];
      first |= (1 << i);
    } else
      ans += c0[i];
  }
  cout << ans << ' ' << first;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
struct tri {
  int first, second, z;
};
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 3e5 + 10;
const long long mod = 1e9 + 7;
const double PI = acos(0) * 2;
int n, arr[40][N], brr[40][N];
int first;
long long c0[40], c1[40], ans;
void dfs(vector<int> &v, int d) {
  if (v.size() <= 1) return;
  vector<int> v0, v1;
  for (auto i : v) {
    if (i & (1 << d)) {
      v1.push_back(i);
      c1[d] += v0.size();
    } else {
      v0.push_back(i);
      c0[d] += v1.size();
    }
  }
  if (d) {
    dfs(v0, d - 1);
    dfs(v1, d - 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  dfs(v, 30);
  for (int i = 0; i <= 30; i++) {
    if (c1[i] < c0[i]) {
      ans += c1[i];
      first |= (1 << i);
    } else
      ans += c0[i];
  }
  cout << ans << ' ' << first;
  return 0;
}

```