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 = 3e5 + 5;
long long f[32][2];
void dp(int b, vector<int> v) {
  if (b == -1 || v.empty()) return;
  int zero = 0, one = 0;
  vector<int> v0, v1;
  for (int x : v) {
    if (x >> b & 1) {
      v1.push_back(x);
      ++one;
      f[b][1] += zero;
    } else {
      v0.push_back(x);
      ++zero;
      f[b][0] += one;
    }
  }
  dp(b - 1, v0);
  dp(b - 1, v1);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  dp(30, a);
  long long inv = 0;
  int x = 0;
  for (int i = 30; i >= 0; --i) {
    if (f[i][1] < f[i][0])
      x |= 1 << i, inv += f[i][1];
    else
      inv += f[i][0];
  }
  printf("%lld %d\n", inv, x);
  return 0;
}
