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;
int a[300000];
long long int m[31][2], x, r;
void s(int d, vector<int> v) {
  if (d < 0 or !v.size()) return;
  int c[2] = {0, 0};
  vector<int> p[2];
  for (int i : v) {
    int b = !!((1 << d) & a[i]);
    c[b]++;
    p[b].push_back(i);
    m[d][b] += c[!b];
  }
  s(d - 1, p[0]);
  s(d - 1, p[1]);
}
int main() {
  int n;
  cin >> n;
  vector<int> ids;
  for (int i = 0; i < n; i++) cin >> a[i], ids.push_back(i);
  s(30, ids);
  for (int i = 0; i < 31; i++)
    r += min(m[i][1], m[i][0]), x += (1 << i) * (m[i][1] < m[i][0]);
  cout << r << " " << x << "\n";
}
