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;
long long x = 1073741824;
long long now = 1;
long long und[31][2], s[10000000], g[10000000][2];
void add(long long y) {
  long long v = 0;
  for (int i = 30; i >= 0; i--) {
    long long bit = (y >> i) % 2;
    if (g[v][bit] == 0) {
      now++;
      g[v][bit] = now;
    }
    if (bit == 1)
      und[i][0] += s[g[v][0]];
    else
      und[i][1] += s[g[v][1]];
    v = g[v][bit];
    s[v]++;
  }
}
int main() {
  ios::sync_with_stdio(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) add(a[i]);
  long long ans = 0, y = 0;
  for (int i = 30; i >= 0; i--) {
    if (und[i][0] < und[i][1]) ans += 1 << i;
    y += min(und[i][0], und[i][1]);
  }
  cout << y << ' ' << ans << "\n";
}
