#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, set<T> const& v) {
  for (auto i = v.begin(); i != v.end(); i++) out << (*i) << ' ';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, multiset<T> const& v) {
  for (auto i = v.begin(); i != v.end(); i++) out << (*i) << ' ';
  return out;
}
template <typename T, typename V>
ostream& operator<<(ostream& out, map<T, V> const& v) {
  for (auto i = v.begin(); i != v.end(); i++)
    out << "\n" << (i->first) << ":" << (i->second);
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 3e6;
vector<vector<int>> trie;
vector<int> cnt;
long long lev0[40];
long long lev1[40];
vector<int> new_node() { return vector<int>(2, -1); }
int solve() {
  for (int i = 0; i < 40; i++) {
    lev0[i] = 0;
    lev1[i] = 0;
  }
  int n;
  cin >> n;
  vector<int> arr(n);
  trie.push_back(new_node());
  cnt.push_back(0);
  int ind = 1;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    int curr = 0;
    for (int j = 29; j >= 0; j--) {
      int bit = bool(arr[i] & (1 << j));
      if (trie[curr][bit] == -1) {
        trie.push_back(new_node());
        cnt.push_back(0);
        trie[curr][bit] = ind;
        ind++;
      }
      if (trie[curr][!bit] == -1) {
        trie.push_back(new_node());
        cnt.push_back(0);
        trie[curr][!bit] = ind;
        ind++;
      }
      if (bit == 0) {
        lev1[j] += cnt[trie[curr][!bit]];
      } else {
        lev0[j] += cnt[trie[curr][!bit]];
      }
      curr = trie[curr][bit];
      cnt[curr]++;
    }
  }
  int ans = 0;
  long long inv = 0;
  for (int i = 0; i < 30; i++) {
    inv += min(lev0[i], lev1[i]);
    if (lev0[i] < lev1[i]) {
      ans += (1 << i);
    }
  }
  cout << inv << " " << ans << "\n";
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) solve();
  return EXIT_SUCCESS;
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, set<T> const& v) {
  for (auto i = v.begin(); i != v.end(); i++) out << (*i) << ' ';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, multiset<T> const& v) {
  for (auto i = v.begin(); i != v.end(); i++) out << (*i) << ' ';
  return out;
}
template <typename T, typename V>
ostream& operator<<(ostream& out, map<T, V> const& v) {
  for (auto i = v.begin(); i != v.end(); i++)
    out << "\n" << (i->first) << ":" << (i->second);
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 3e6;
vector<vector<int>> trie;
vector<int> cnt;
long long lev0[40];
long long lev1[40];
vector<int> new_node() { return vector<int>(2, -1); }
int solve() {
  for (int i = 0; i < 40; i++) {
    lev0[i] = 0;
    lev1[i] = 0;
  }
  int n;
  cin >> n;
  vector<int> arr(n);
  trie.push_back(new_node());
  cnt.push_back(0);
  int ind = 1;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    int curr = 0;
    for (int j = 29; j >= 0; j--) {
      int bit = bool(arr[i] & (1 << j));
      if (trie[curr][bit] == -1) {
        trie.push_back(new_node());
        cnt.push_back(0);
        trie[curr][bit] = ind;
        ind++;
      }
      if (trie[curr][!bit] == -1) {
        trie.push_back(new_node());
        cnt.push_back(0);
        trie[curr][!bit] = ind;
        ind++;
      }
      if (bit == 0) {
        lev1[j] += cnt[trie[curr][!bit]];
      } else {
        lev0[j] += cnt[trie[curr][!bit]];
      }
      curr = trie[curr][bit];
      cnt[curr]++;
    }
  }
  int ans = 0;
  long long inv = 0;
  for (int i = 0; i < 30; i++) {
    inv += min(lev0[i], lev1[i]);
    if (lev0[i] < lev1[i]) {
      ans += (1 << i);
    }
  }
  cout << inv << " " << ans << "\n";
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) solve();
  return EXIT_SUCCESS;
}

```