#include <bits/stdc++.h>
using namespace std;
void see(int x) { cout << x; }
void see(long long int x) { cout << x; }
void see(long double x) { cout << x; }
void see(char x) { cout << '\'' << x << '\''; }
void see(string x) { cout << '\"' << x << '\"'; }
template <typename T, typename V>
void see(const pair<T, V> &x) {
  cout << '{';
  see(x.first);
  cout << ',';
  see(x.second);
  cout << '}';
}
template <typename T>
void see(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), see(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  see(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long int gcd(long long int a, long long int b) {
  return (!b) ? a : gcd(b, a % b);
}
const int N = 30 * 3e5 + 5;
int val[N];
int trie[N][2];
long long int dp[32][2], cur;
void add(int x) {
  int st = 0;
  for (int i = 31; i >= 0; i--) {
    int bt = ((x >> i) & 1);
    if (!trie[st][bt]) trie[st][bt] = ++cur;
    dp[i][bt] += val[trie[st][!bt]];
    st = trie[st][bt];
    val[st] += 1;
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  long long int inv = 0;
  int ans = 0;
  for (int i = 0; i <= 31; i++) {
    if (dp[i][1] < dp[i][0]) {
      inv += 1ll * dp[i][1];
      ans += (1 << i);
    } else {
      inv += 1ll * dp[i][0];
    }
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

### Prompt

In Cpp, your task is to solve 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;
void see(int x) { cout << x; }
void see(long long int x) { cout << x; }
void see(long double x) { cout << x; }
void see(char x) { cout << '\'' << x << '\''; }
void see(string x) { cout << '\"' << x << '\"'; }
template <typename T, typename V>
void see(const pair<T, V> &x) {
  cout << '{';
  see(x.first);
  cout << ',';
  see(x.second);
  cout << '}';
}
template <typename T>
void see(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), see(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  see(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long int gcd(long long int a, long long int b) {
  return (!b) ? a : gcd(b, a % b);
}
const int N = 30 * 3e5 + 5;
int val[N];
int trie[N][2];
long long int dp[32][2], cur;
void add(int x) {
  int st = 0;
  for (int i = 31; i >= 0; i--) {
    int bt = ((x >> i) & 1);
    if (!trie[st][bt]) trie[st][bt] = ++cur;
    dp[i][bt] += val[trie[st][!bt]];
    st = trie[st][bt];
    val[st] += 1;
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  long long int inv = 0;
  int ans = 0;
  for (int i = 0; i <= 31; i++) {
    if (dp[i][1] < dp[i][0]) {
      inv += 1ll * dp[i][1];
      ans += (1 << i);
    } else {
      inv += 1ll * dp[i][0];
    }
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

```