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 int ans[2][32];
void inversions(vector<long long int> &v, vector<long long int> &index,
                long long int b) {
  if (index.size() <= 1 || b < 0) {
    return;
  }
  vector<long long int> v1;
  vector<long long int> v2;
  long long int cnt = 0;
  for (long long int i = 0; i < index.size(); i++) {
    if (v[index[i]] >> b & 1) {
      cnt++;
    } else {
      ans[0][b] += cnt;
    }
  }
  cnt = 0;
  for (long long int i = 0; i < index.size(); i++) {
    if (v[index[i]] >> b & 1) {
      ans[1][b] += cnt;
      v1.push_back(index[i]);
    } else {
      cnt++;
      v2.push_back(index[i]);
    }
  }
  inversions(v, v2, b - 1);
  inversions(v, v1, b - 1);
}
int main() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  vector<long long int> index(n);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    index[i] = i;
  }
  inversions(v, index, 30);
  long long int a = 0, num = 0;
  long long int fac = (1 << 30);
  for (long long int i = 30; i >= 0; i--) {
    if (ans[0][i] > ans[1][i]) {
      a = a + ans[1][i];
      num = num + fac;
    } else {
      a = a + ans[0][i];
    }
    fac = fac / 2;
  }
  cout << a << " " << num << "\n";
  return 0;
}
