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 mxN = 1e5 + 7;
const int inf = 1e9 + 7;
const int mod = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<long long> zero(31, 0), one(31, 0);
  function<void(vector<int> &, int)> rec = [&](vector<int> &t, int b) {
    int m = t.size();
    if (b < 0 || m == 0) return;
    vector<int> z, o;
    int x = 0, y = 0;
    for (int i = 0; i < m; ++i) {
      if ((t[i] >> b) & 1) {
        ++y;
        one[b] += x;
        o.push_back(t[i]);
      } else {
        ++x;
        zero[b] += y;
        z.push_back(t[i]);
      }
    }
    rec(z, b - 1);
    rec(o, b - 1);
  };
  rec(a, 30);
  long long inv = 0, num = 0;
  for (int i = 30; i >= 0; --i) {
    inv += min(zero[i], one[i]);
    num |= (zero[i] <= one[i] ? 0 : (1 << i));
  }
  cout << inv << ' ' << num << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  for (int tt = 1; tt <= T; ++tt) {
    solve();
  }
  return 0;
}
