#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
struct Trie {
  vector<pair<int, int> > ind;
  int key;
};
int main() {
  long long n, x = 0, mxbit = -1, res = 0;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  while ((1 << (mxbit + 1)) <= a.back().first) mxbit++;
  if (mxbit == -1) {
    cout << "0 0\n";
    return 0;
  }
  Trie* root = new Trie();
  root->ind = a;
  root->key = 0;
  vector<Trie*> cur_level;
  cur_level.push_back(root);
  while (!cur_level.empty() && mxbit >= 0) {
    vector<Trie*> temp;
    long long total_sum = 0, sum = 0;
    for (Trie* cur : cur_level) {
      pair<int, int> p = {(1 << mxbit) + cur->key, 0};
      int pivot =
          lower_bound(cur->ind.begin(), cur->ind.end(), p) - cur->ind.begin();
      vector<pair<int, int> > first_half, second_half;
      vector<int> fi, si;
      for (int i = 0; i < cur->ind.size(); i++) {
        if (i < pivot) {
          first_half.push_back(cur->ind[i]);
          fi.push_back(cur->ind[i].second);
        } else {
          second_half.push_back(cur->ind[i]);
          si.push_back(cur->ind[i].second);
        }
      }
      if (first_half.size() > 0) {
        Trie* t = new Trie();
        t->ind = first_half;
        t->key = cur->key;
        temp.push_back(t);
      }
      if (second_half.size() > 0) {
        Trie* tt = new Trie();
        tt->ind = second_half;
        tt->key = cur->key + (1 << mxbit);
        temp.push_back(tt);
      }
      sort(fi.begin(), fi.end());
      sort(si.begin(), si.end());
      total_sum += (long long)fi.size() * (long long)si.size();
      long long p1 = 0, p2 = 0, cnt = 0, tmp_sum = 0;
      while (p1 < fi.size() && p2 < si.size()) {
        if (fi[p1] > si[p2]) {
          p2++;
          cnt++;
        } else {
          tmp_sum += cnt;
          p1++;
        }
      }
      if (p2 == si.size()) {
        tmp_sum += (long long)(fi.size() - p1) * cnt;
      }
      sum += tmp_sum;
    }
    cur_level = temp;
    if (sum > total_sum / 2) {
      x += (1 << mxbit);
      sum = total_sum - sum;
    }
    res += sum;
    mxbit--;
  }
  cout << res << " " << x << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 long long MOD = 1e9 + 7;
struct Trie {
  vector<pair<int, int> > ind;
  int key;
};
int main() {
  long long n, x = 0, mxbit = -1, res = 0;
  cin >> n;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  while ((1 << (mxbit + 1)) <= a.back().first) mxbit++;
  if (mxbit == -1) {
    cout << "0 0\n";
    return 0;
  }
  Trie* root = new Trie();
  root->ind = a;
  root->key = 0;
  vector<Trie*> cur_level;
  cur_level.push_back(root);
  while (!cur_level.empty() && mxbit >= 0) {
    vector<Trie*> temp;
    long long total_sum = 0, sum = 0;
    for (Trie* cur : cur_level) {
      pair<int, int> p = {(1 << mxbit) + cur->key, 0};
      int pivot =
          lower_bound(cur->ind.begin(), cur->ind.end(), p) - cur->ind.begin();
      vector<pair<int, int> > first_half, second_half;
      vector<int> fi, si;
      for (int i = 0; i < cur->ind.size(); i++) {
        if (i < pivot) {
          first_half.push_back(cur->ind[i]);
          fi.push_back(cur->ind[i].second);
        } else {
          second_half.push_back(cur->ind[i]);
          si.push_back(cur->ind[i].second);
        }
      }
      if (first_half.size() > 0) {
        Trie* t = new Trie();
        t->ind = first_half;
        t->key = cur->key;
        temp.push_back(t);
      }
      if (second_half.size() > 0) {
        Trie* tt = new Trie();
        tt->ind = second_half;
        tt->key = cur->key + (1 << mxbit);
        temp.push_back(tt);
      }
      sort(fi.begin(), fi.end());
      sort(si.begin(), si.end());
      total_sum += (long long)fi.size() * (long long)si.size();
      long long p1 = 0, p2 = 0, cnt = 0, tmp_sum = 0;
      while (p1 < fi.size() && p2 < si.size()) {
        if (fi[p1] > si[p2]) {
          p2++;
          cnt++;
        } else {
          tmp_sum += cnt;
          p1++;
        }
      }
      if (p2 == si.size()) {
        tmp_sum += (long long)(fi.size() - p1) * cnt;
      }
      sum += tmp_sum;
    }
    cur_level = temp;
    if (sum > total_sum / 2) {
      x += (1 << mxbit);
      sum = total_sum - sum;
    }
    res += sum;
    mxbit--;
  }
  cout << res << " " << x << "\n";
  return 0;
}

```