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 long long inf = 0x3f3f3f3f;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> nums(n);
  for (long long i = 0; i < n; i++) cin >> nums[i];
  vector<vector<long long> > gs;
  gs.push_back(nums);
  long long ans = 0, ansi = 0;
  for (long long b = 31; b >= 0; b--) {
    vector<vector<long long> > ngs;
    long long oi = 0, zi = 0;
    long long bi = 1 << b;
    for (auto ns : gs) {
      long long o = 0, z = 0;
      vector<long long> ones, zeros;
      for (auto x : ns) {
        if (x & bi) {
          o++;
          oi += z;
          ones.push_back(x);
        } else {
          z++;
          zi += o;
          zeros.push_back(x);
        }
      }
      if (!ones.empty()) ngs.push_back(ones);
      if (!zeros.empty()) ngs.push_back(zeros);
    }
    gs = ngs;
    if (oi >= zi) {
      ansi += zi;
    } else {
      ans |= bi;
      ansi += oi;
    }
  }
  cout << ansi << ' ' << ans << endl;
}
