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;
const int MAXN = 300300, TRIE = 30 * MAXN + 10;
int trie[2][TRIE];
long long int freq[TRIE], inversions[TRIE];
int curNode = 1;
void insert(int x) {
  int node = 0;
  for (int bit = 29; bit >= 0; bit--) {
    int e = (1 << bit) & x;
    if (e) e = 1;
    if (!trie[e][node]) trie[e][node] = curNode++;
    if (e && trie[e ^ 1][node]) {
      inversions[node] += freq[trie[e ^ 1][node]];
    }
    node = trie[e][node];
    freq[node]++;
  }
}
long long int totalPairs[30], sumInv[30];
void dfs(int node, int depth) {
  long long int left = 0, right = 0;
  int L = trie[0][node], R = trie[1][node];
  if (L) left = freq[L];
  if (R) right = freq[R];
  totalPairs[depth] += left * right;
  sumInv[depth] += inversions[node];
  if (L) dfs(L, depth - 1);
  if (R) dfs(R, depth - 1);
}
void sol() {
  int n;
  cin >> n;
  vector<int> vals(n);
  for (int& x : vals) cin >> x;
  for (int i = n - 1; i >= 0; i--) {
    insert(vals[i]);
  }
  dfs(0, 29);
  int res = 0;
  long long int sum = 0;
  for (int i = 29; i >= 0; i--) {
    long long int other = totalPairs[i] - sumInv[i];
    if (other < sumInv[i]) {
      res ^= (1 << i);
    }
    sum += min(other, sumInv[i]);
  }
  cout << sum << " " << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  sol();
  return 0;
}
