#include <bits/stdc++.h>
using namespace std;
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << " : " << m.second << ") ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " |";
  __f(comma + 1, args...);
}
vector<pair<long long, long long>> invs(30);
void calc(vector<int>& arr, int b) {
  if (b < 0 or arr.size() < 2) return;
  vector<int> arr0, arr1;
  long long sm = 0, rev = 0;
  int cnt0 = 0, cnt1 = 0;
  for (int a : arr) {
    if (a & (1 << b)) {
      rev += cnt0;
      cnt1++;
      arr1.push_back(a);
    } else {
      sm += cnt1;
      cnt0++;
      arr0.push_back(a);
    }
  }
  invs[b].first += sm;
  invs[b].second += rev;
  calc(arr0, b - 1);
  calc(arr1, b - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int& i : arr) cin >> i;
  calc(arr, 29);
  long long inv = 0, ans = 0;
  for (int i = 0; i < 30; ++i) {
    inv += min(invs[i].first, invs[i].second);
    if (invs[i].second < invs[i].first) ans += (1 << i);
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1, tc = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << " : " << m.second << ") ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " |";
  __f(comma + 1, args...);
}
vector<pair<long long, long long>> invs(30);
void calc(vector<int>& arr, int b) {
  if (b < 0 or arr.size() < 2) return;
  vector<int> arr0, arr1;
  long long sm = 0, rev = 0;
  int cnt0 = 0, cnt1 = 0;
  for (int a : arr) {
    if (a & (1 << b)) {
      rev += cnt0;
      cnt1++;
      arr1.push_back(a);
    } else {
      sm += cnt1;
      cnt0++;
      arr0.push_back(a);
    }
  }
  invs[b].first += sm;
  invs[b].second += rev;
  calc(arr0, b - 1);
  calc(arr1, b - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int& i : arr) cin >> i;
  calc(arr, 29);
  long long inv = 0, ans = 0;
  for (int i = 0; i < 30; ++i) {
    inv += min(invs[i].first, invs[i].second);
    if (invs[i].second < invs[i].first) ans += (1 << i);
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1, tc = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```