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;
using ll = long long;
const int N = 3e5 + 10;
const int inf = 1e9;
int n, arr[N];
ll f[35][3];
void solve(int bit, vector<int> v) {
  if (bit < 0 || v.empty()) return;
  vector<int> v1, v2;
  ll ones = 0, zeros = 0;
  for (int x : v) {
    if (x & (1 << bit)) {
      ones++;
      f[bit][1] += zeros;
      v1.push_back(x);
    } else {
      zeros++;
      f[bit][0] += ones;
      v2.push_back(x);
    }
  }
  solve(bit - 1, v1), solve(bit - 1, v2);
}
int main() {
  cin >> n;
  vector<int> v(n);
  for (int I = (0); I <= (n - 1); I++) cin >> v[I];
  solve(31, v);
  ll ans = 0, inv = 0;
  for (ll i = 0; i < 32; i++) {
    if (f[i][1] < f[i][0]) {
      ans |= (1LL << i);
      inv += (ll)f[i][1];
    } else
      inv += (ll)f[i][0];
  }
  printf("%lld %lld\n", inv, ans);
  return 0;
}
