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 N = 5e6 + 5;
int trie[N][2], cnt = 0;
long long dp[N][2];
vector<int> node[N];
void add(int x, int id) {
  int cur = 0;
  for (int i = 29; i >= 0; i--) {
    int c = (x >> i) & 1;
    if (!trie[cur][c]) trie[cur][c] = ++cnt;
    cur = trie[cur][c];
    node[cur].push_back(id);
  }
}
void go(int v, int b = 29) {
  int l = trie[v][0], r = trie[v][1];
  if (l > 0) go(l, b - 1);
  if (r > 0) go(r, b - 1);
  if (l == 0 || r == 0) return;
  long long ans = 0;
  int ptr = 0;
  for (int pos : node[l]) {
    while (ptr < (int)node[r].size() && node[r][ptr] < pos) ptr++;
    ans += ptr;
  }
  dp[b][0] += ans;
  dp[b][1] += (long long)node[l].size() * (long long)node[r].size() - ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    add(a, i);
  }
  go(0);
  long long inv = 0;
  int res = 0;
  for (int i = 0; i < 30; i++) {
    inv += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) res += (1 << i);
  }
  cout << inv << ' ' << res << '\n';
}
