#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const long long MOD = 1000000007;
const int max_n = 3e5 + 100;
int a[max_n];
int n;
long long ans0 = 0;
long long ans = 0;
void Solve(int x, vector<vector<int>>& vc) {
  if (x < 0) return;
  vector<vector<int>> vn;
  long long res1 = 0, res2 = 0;
  for (int i = 0; i < vc.size(); ++i) {
    vector<int> v1, v2;
    long long cnt1 = 0, cnt2 = 0;
    for (int num : vc[i]) {
      if ((num >> x) & 1) {
        ++cnt1;
        res1 += cnt2;
        v1.push_back(num);
      } else {
        ++cnt2;
        res2 += cnt1;
        v2.push_back(num);
      }
    }
    if (!v1.empty()) vn.push_back(v1);
    if (!v2.empty()) vn.push_back(v2);
  }
  if (res1 < res2) {
    ans0 |= (1 << x);
    ans += res1;
  } else {
    ans += res2;
  }
  if (!vn.empty()) Solve(x - 1, vn);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  vector<vector<int>> vc;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    int num;
    cin >> num;
    v.push_back(num);
  }
  vc.push_back(v);
  Solve(31, vc);
  cout << ans << " " << ans0 << endl;
}

### Prompt

Generate 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;
const int N = 200005;
const long long MOD = 1000000007;
const int max_n = 3e5 + 100;
int a[max_n];
int n;
long long ans0 = 0;
long long ans = 0;
void Solve(int x, vector<vector<int>>& vc) {
  if (x < 0) return;
  vector<vector<int>> vn;
  long long res1 = 0, res2 = 0;
  for (int i = 0; i < vc.size(); ++i) {
    vector<int> v1, v2;
    long long cnt1 = 0, cnt2 = 0;
    for (int num : vc[i]) {
      if ((num >> x) & 1) {
        ++cnt1;
        res1 += cnt2;
        v1.push_back(num);
      } else {
        ++cnt2;
        res2 += cnt1;
        v2.push_back(num);
      }
    }
    if (!v1.empty()) vn.push_back(v1);
    if (!v2.empty()) vn.push_back(v2);
  }
  if (res1 < res2) {
    ans0 |= (1 << x);
    ans += res1;
  } else {
    ans += res2;
  }
  if (!vn.empty()) Solve(x - 1, vn);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  vector<vector<int>> vc;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    int num;
    cin >> num;
    v.push_back(num);
  }
  vc.push_back(v);
  Solve(31, vc);
  cout << ans << " " << ans0 << endl;
}

```