#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> a(1, vector<int>(n));
  for (int i = 0; i < n; i++) cin >> a[0][i];
  long long ansx = 0, ansinv = 0;
  for (int pow = 29; pow >= 0; pow--) {
    vector<vector<int>> b;
    int mod = (1 << pow);
    long long w0 = 0, w1 = 0;
    for (auto &arr : a) {
      vector<int> a0, a1;
      vector<int> cnt(arr.size());
      for (int i = 0; i < (int)arr.size(); i++) {
        bool flag = (arr[i] & mod) > 0;
        if (flag) {
          a1.push_back(arr[i]);
          cnt[i] = 1;
        } else {
          a0.push_back(arr[i]);
          cnt[i] = 0;
        }
      }
      if (!a0.empty()) b.push_back(a0);
      if (!a1.empty()) b.push_back(a1);
      vector<int> tmp(arr.size());
      tmp[0] = cnt[0];
      for (int i = 1; i < (int)tmp.size(); i++) tmp[i] = tmp[i - 1] + cnt[i];
      for (int i = 0; i < (int)tmp.size(); i++)
        if (!cnt[i]) w0 += tmp[i];
      tmp[0] = 1 - cnt[0];
      for (int i = 1; i < (int)tmp.size(); i++)
        tmp[i] = tmp[i - 1] + (1 - cnt[i]);
      for (int i = 0; i < (int)tmp.size(); i++)
        if (cnt[i]) w1 += tmp[i];
    }
    if (w0 <= w1) {
      ansinv += w0;
    } else {
      ansinv += w1;
      ansx += mod;
    }
    swap(a, b);
  }
  cout << ansinv << ' ' << ansx;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> a(1, vector<int>(n));
  for (int i = 0; i < n; i++) cin >> a[0][i];
  long long ansx = 0, ansinv = 0;
  for (int pow = 29; pow >= 0; pow--) {
    vector<vector<int>> b;
    int mod = (1 << pow);
    long long w0 = 0, w1 = 0;
    for (auto &arr : a) {
      vector<int> a0, a1;
      vector<int> cnt(arr.size());
      for (int i = 0; i < (int)arr.size(); i++) {
        bool flag = (arr[i] & mod) > 0;
        if (flag) {
          a1.push_back(arr[i]);
          cnt[i] = 1;
        } else {
          a0.push_back(arr[i]);
          cnt[i] = 0;
        }
      }
      if (!a0.empty()) b.push_back(a0);
      if (!a1.empty()) b.push_back(a1);
      vector<int> tmp(arr.size());
      tmp[0] = cnt[0];
      for (int i = 1; i < (int)tmp.size(); i++) tmp[i] = tmp[i - 1] + cnt[i];
      for (int i = 0; i < (int)tmp.size(); i++)
        if (!cnt[i]) w0 += tmp[i];
      tmp[0] = 1 - cnt[0];
      for (int i = 1; i < (int)tmp.size(); i++)
        tmp[i] = tmp[i - 1] + (1 - cnt[i]);
      for (int i = 0; i < (int)tmp.size(); i++)
        if (cnt[i]) w1 += tmp[i];
    }
    if (w0 <= w1) {
      ansinv += w0;
    } else {
      ansinv += w1;
      ansx += mod;
    }
    swap(a, b);
  }
  cout << ansinv << ' ' << ansx;
  return 0;
}

```