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 int64_t INF = (int64_t)1e9 + 7;
int64_t n;
int64_t invcnt[2][32];
void update(vector<int64_t>& a, const int64_t digit) {
  if (digit < 0 || a.empty()) return;
  vector<int64_t> zeros, ones;
  int64_t cnt0 = 0, cnt1 = 0;
  int64_t inv0 = 0, inv1 = 0;
  for (int64_t i = 0; i < (int64_t)a.size(); i++) {
    if (a[i] & (1 << digit)) {
      cnt1++;
      inv1 += cnt0;
      ones.push_back(a[i]);
    } else {
      cnt0++;
      inv0 += cnt1;
      zeros.push_back(a[i]);
    }
  }
  invcnt[0][digit] += inv0;
  invcnt[1][digit] += inv1;
  update(zeros, digit - 1);
  update(ones, digit - 1);
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  memset(invcnt, 0, sizeof(invcnt));
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t& x : a) cin >> x;
  update(a, 30);
  int64_t x = 0, inv = 0;
  for (int64_t i = 30; i >= 0; i--) {
    if (invcnt[0][i] <= invcnt[1][i]) {
      inv += invcnt[0][i];
    } else {
      inv += invcnt[1][i];
      x |= (1 << i);
    }
  }
  cout << inv << " " << x << endl;
  return 0;
}
