#include <bits/stdc++.h>
using namespace std;
using vi = vector<long long>;
using vs = vector<string>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ii = pair<int, int>;
using iii = pair<ii, ii>;
using vii = vector<ii>;
using viii = vector<pair<ii, ii>>;
using vvii = vector<vii>;
using vs = vector<string>;
using vd = vector<double>;
using vvd = vector<vd>;
using viid = vector<pair<ii, double>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vvi separats(1, vi(n));
  for (auto& x : separats[0]) cin >> x;
  long long sol = 0;
  long long num = 0;
  for (int it = 30; it >= 0; it--) {
    int act = 1 << it;
    vvi auxs = separats;
    vvi next;
    int granAct = 0;
    for (auto& x : auxs) {
      next.push_back(vi());
      next.push_back(vi());
      for (auto& y : x) {
        if (y & act) {
          next[granAct].push_back(y);
        } else {
          next[granAct + 1].push_back(y);
        }
        y = y & act;
      }
      if (next[granAct].size() == 0) {
        next[granAct] = next[granAct + 1];
        next.pop_back();
        granAct += 1;
      } else if (next[granAct + 1].size() == 0) {
        next.pop_back();
        granAct += 1;
      } else
        granAct += 2;
    }
    long long mal1 = 0, mal0 = 0;
    for (auto& x : auxs) {
      int uns = 0, zeros = 0;
      for (auto& y : x) {
        if (y) {
          mal1 += zeros;
          uns++;
        } else {
          mal0 += uns;
          zeros++;
        }
      }
    }
    if (mal1 < mal0) {
      sol += mal1;
      num += act;
    } else {
      sol += mal0;
    }
    separats = next;
  }
  cout << sol << " " << num << '\n';
}

### Prompt

Please create a solution in cpp 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;
using vi = vector<long long>;
using vs = vector<string>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ii = pair<int, int>;
using iii = pair<ii, ii>;
using vii = vector<ii>;
using viii = vector<pair<ii, ii>>;
using vvii = vector<vii>;
using vs = vector<string>;
using vd = vector<double>;
using vvd = vector<vd>;
using viid = vector<pair<ii, double>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vvi separats(1, vi(n));
  for (auto& x : separats[0]) cin >> x;
  long long sol = 0;
  long long num = 0;
  for (int it = 30; it >= 0; it--) {
    int act = 1 << it;
    vvi auxs = separats;
    vvi next;
    int granAct = 0;
    for (auto& x : auxs) {
      next.push_back(vi());
      next.push_back(vi());
      for (auto& y : x) {
        if (y & act) {
          next[granAct].push_back(y);
        } else {
          next[granAct + 1].push_back(y);
        }
        y = y & act;
      }
      if (next[granAct].size() == 0) {
        next[granAct] = next[granAct + 1];
        next.pop_back();
        granAct += 1;
      } else if (next[granAct + 1].size() == 0) {
        next.pop_back();
        granAct += 1;
      } else
        granAct += 2;
    }
    long long mal1 = 0, mal0 = 0;
    for (auto& x : auxs) {
      int uns = 0, zeros = 0;
      for (auto& y : x) {
        if (y) {
          mal1 += zeros;
          uns++;
        } else {
          mal0 += uns;
          zeros++;
        }
      }
    }
    if (mal1 < mal0) {
      sol += mal1;
      num += act;
    } else {
      sol += mal0;
    }
    separats = next;
  }
  cout << sol << " " << num << '\n';
}

```