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 power[35];
pair<long long int, long long int> vals[31];
void func(vector<long long int> &v, long long int z) {
  if (v.size() == 0) {
    return;
  }
  if (z == -1) {
    return;
  }
  vector<long long int> v1;
  vector<long long int> v2;
  long long int c0 = 0;
  long long int c1 = 0;
  long long int a0 = 0;
  long long int a1 = 0;
  for (long long int i = 0; i < v.size(); i++) {
    if ((v[i] & power[z]) == 0) {
      c0++;
      a0 += c1;
      v1.push_back(v[i]);
    } else {
      c1++;
      a1 += c0;
      v2.push_back(v[i]);
    }
  }
  vals[z].first += a0;
  vals[z].second += a1;
  func(v1, z - 1);
  func(v2, z - 1);
  return;
}
signed main() {
  power[0] = 1;
  for (long long int i = 1; i <= 30; i++) {
    power[i] = power[i - 1] * 2;
  }
  long long int n;
  cin >> n;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    long long int z;
    cin >> z;
    v.push_back(z);
  }
  func(v, 30);
  long long int ans = 0;
  long long int inv = 0;
  for (long long int i = 30; i >= 0; i--) {
    inv += min(vals[i].second, vals[i].first);
    if (vals[i].first <= vals[i].second) {
      ;
    } else {
      ans += power[i];
    }
  }
  cout << inv << " " << ans;
}
