#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
mt19937 mt(736);
ll cnt(const vector<bool> &arr) {
  int ones = 0;
  ll ans = 0;
  for (auto it : arr)
    if (it)
      ones++;
    else
      ans += ones;
  return ans;
}
void slv(const vector<unsigned> &arr, vector<pair<ll, ll>> &q, int bit) {
  if (bit < 0 || arr.size() < 2) return;
  auto x = 1u << (unsigned)bit;
  vector<bool> tmp(arr.size()), ntmp(arr.size());
  for (int j = 0; j < tmp.size(); j++) ntmp[j] = !(tmp[j] = (arr[j] & x) > 0);
  q[bit].first += cnt(tmp);
  q[bit].second += cnt(ntmp);
  vector<unsigned> ze, on;
  for (auto it : arr) (it & x ? on : ze).push_back(it);
  slv(ze, q, bit - 1);
  slv(on, q, bit - 1);
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  vector<unsigned> arr(n);
  for (auto &it : arr) cin >> it;
  vector<pair<ll, ll>> q(31);
  slv(arr, q, (int)q.size() - 1);
  ll ans = 0;
  unsigned x = 0;
  for (unsigned i = 0; i < q.size(); i++) {
    if (q[i].second < q[i].first) {
      x |= (1u << i);
    }
    ans += min(q[i].first, q[i].second);
  }
  cout << ans << ' ' << x << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

### 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 ll = long long;
using ull = unsigned long long;
mt19937 mt(736);
ll cnt(const vector<bool> &arr) {
  int ones = 0;
  ll ans = 0;
  for (auto it : arr)
    if (it)
      ones++;
    else
      ans += ones;
  return ans;
}
void slv(const vector<unsigned> &arr, vector<pair<ll, ll>> &q, int bit) {
  if (bit < 0 || arr.size() < 2) return;
  auto x = 1u << (unsigned)bit;
  vector<bool> tmp(arr.size()), ntmp(arr.size());
  for (int j = 0; j < tmp.size(); j++) ntmp[j] = !(tmp[j] = (arr[j] & x) > 0);
  q[bit].first += cnt(tmp);
  q[bit].second += cnt(ntmp);
  vector<unsigned> ze, on;
  for (auto it : arr) (it & x ? on : ze).push_back(it);
  slv(ze, q, bit - 1);
  slv(on, q, bit - 1);
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n;
  cin >> n;
  vector<unsigned> arr(n);
  for (auto &it : arr) cin >> it;
  vector<pair<ll, ll>> q(31);
  slv(arr, q, (int)q.size() - 1);
  ll ans = 0;
  unsigned x = 0;
  for (unsigned i = 0; i < q.size(); i++) {
    if (q[i].second < q[i].first) {
      x |= (1u << i);
    }
    ans += min(q[i].first, q[i].second);
  }
  cout << ans << ' ' << x << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}

```