#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
template <typename T>
ll merge_sort(vector<T>& v, int l, int r) {
  if (r <= l) return 0LL;
  static vector<T> u;
  int m = l + (r - l) / 2;
  ll res = merge_sort(v, l, m) + merge_sort(v, m + 1, r);
  int x = l, y = m + 1;
  while (x <= m && y <= r) {
    if (v[x] <= v[y])
      u.push_back(v[x++]);
    else
      u.push_back(v[y++]), res += m - x + 1;
  }
  while (x <= m) u.push_back(v[x++]);
  while (y <= r) u.push_back(v[y++]);
  while (l <= r) v[r] = u[r - l], --r;
  u.clear();
  return res;
}
template <typename T>
ll inversions(vector<T> v) {
  return merge_sort(v, 0, (int)v.size() - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n, 0);
  int logmax = 0;
  for (auto& z : a) {
    cin >> z;
    if (z) logmax = max(logmax, 31 - __builtin_clz(z));
  }
  int x = 0;
  ll last = inversions(a);
  for (int i = logmax; i >= 0; --i) {
    for (auto& z : a) z ^= 1 << i;
    ll cur = inversions(a);
    if (last <= cur)
      for (auto& z : a) z ^= 1 << i;
    else
      x ^= 1 << i, last = cur;
  }
  cout << last << " " << x << '\n';
  exit(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;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
template <typename T>
ll merge_sort(vector<T>& v, int l, int r) {
  if (r <= l) return 0LL;
  static vector<T> u;
  int m = l + (r - l) / 2;
  ll res = merge_sort(v, l, m) + merge_sort(v, m + 1, r);
  int x = l, y = m + 1;
  while (x <= m && y <= r) {
    if (v[x] <= v[y])
      u.push_back(v[x++]);
    else
      u.push_back(v[y++]), res += m - x + 1;
  }
  while (x <= m) u.push_back(v[x++]);
  while (y <= r) u.push_back(v[y++]);
  while (l <= r) v[r] = u[r - l], --r;
  u.clear();
  return res;
}
template <typename T>
ll inversions(vector<T> v) {
  return merge_sort(v, 0, (int)v.size() - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n, 0);
  int logmax = 0;
  for (auto& z : a) {
    cin >> z;
    if (z) logmax = max(logmax, 31 - __builtin_clz(z));
  }
  int x = 0;
  ll last = inversions(a);
  for (int i = logmax; i >= 0; --i) {
    for (auto& z : a) z ^= 1 << i;
    ll cur = inversions(a);
    if (last <= cur)
      for (auto& z : a) z ^= 1 << i;
    else
      x ^= 1 << i, last = cur;
  }
  cout << last << " " << x << '\n';
  exit(0);
}

```