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 power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
pair<long long, long long> DivideNConq(vector<vector<int> > v, int bit) {
  if (bit < 0) return {0, 0};
  vector<vector<int> > R;
  long long bit_pos0 = 0, bit_pos1 = 0;
  for (auto it : v) {
    vector<int> v0, v1;
    long long ones = 0, zeros = 0;
    for (auto i : it) {
      if ((i >> bit) & 1) {
        ones++;
        bit_pos1 += zeros;
        v1.push_back(i);
      } else {
        zeros++;
        bit_pos0 += ones;
        v0.push_back(i);
      }
    }
    if (v0.size() > 0) R.push_back(v0);
    if (v1.size() > 0) R.push_back(v1);
  }
  auto p = DivideNConq(R, bit - 1);
  int x = (bit_pos1 < bit_pos0);
  return {min(bit_pos0, bit_pos1) + p.first, power(2, bit) * x + p.second};
}
void nikal_pehli_fursat_mai() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  auto ans = DivideNConq({a}, 30);
  cout << ans.first << " " << ans.second << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    nikal_pehli_fursat_mai();
  }
}
