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.
#include <bits/stdc++.h>
using namespace std;
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...);
}
long long int dp[32][2];
class node {
 public:
  node* left = NULL;
  node* right = NULL;
  long long int cnt = 0;
};
class trie {
  node* root;

 public:
  trie() { root = new node(); }
  void insert(long long int n) {
    node* temp = root;
    for (long long int i = 30; i >= 0; i--) {
      long long int bit = (n >> i) & 1;
      if (bit == 0) {
        if (temp->right != NULL) {
          dp[i][0] += temp->right->cnt;
        }
        if (temp->left == NULL) {
          temp->left = new node();
        }
        temp = temp->left;
        temp->cnt++;
      } else {
        if (temp->left != NULL) {
          dp[i][1] += temp->left->cnt;
        }
        if (temp->right == NULL) {
          temp->right = new node();
        }
        temp = temp->right;
        temp->cnt++;
      }
    }
  }
};
void solve() {
  long long int n;
  cin >> n;
  trie T;
  long long int a;
  for (long long int i = 0; i < n; i++) {
    cin >> a;
    T.insert(a);
  }
  long long int ans = 0;
  long long int x = 0;
  for (long long int i = 30; i >= 0; i--) {
    ans += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) {
      x |= (1LL << i);
    }
  }
  cout << ans << " " << x << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
