#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > cur, tmp;
int main() {
  ios::sync_with_stdio(0);
  int N;
  cin >> N;
  vector<int> pot;
  for (int i = 1; i <= N; ++i) {
    int A;
    cin >> A;
    pot.push_back(A);
  }
  cur.push_back(pot);
  long long ans = 0, x = 0;
  for (int i = 30; ~i; --i) {
    long long cnt1 = 0, cnt0 = 0, sum1 = 0, sum0 = 0;
    tmp.clear();
    for (auto t : cur) {
      cnt1 = cnt0 = 0;
      vector<int> pl, pr;
      for (auto p : t)
        if (p >> i & 1) {
          sum1 += cnt0;
          ++cnt1;
          pl.push_back(p);
        } else {
          sum0 += cnt1;
          ++cnt0;
          pr.push_back(p);
        }
      if (pl.size()) tmp.push_back(pl);
      if (pr.size()) tmp.push_back(pr);
    }
    ans += min(sum0, sum1);
    x |= (sum1 < sum0) << i;
    cur = tmp;
  }
  cout << ans << ' ' << x << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > cur, tmp;
int main() {
  ios::sync_with_stdio(0);
  int N;
  cin >> N;
  vector<int> pot;
  for (int i = 1; i <= N; ++i) {
    int A;
    cin >> A;
    pot.push_back(A);
  }
  cur.push_back(pot);
  long long ans = 0, x = 0;
  for (int i = 30; ~i; --i) {
    long long cnt1 = 0, cnt0 = 0, sum1 = 0, sum0 = 0;
    tmp.clear();
    for (auto t : cur) {
      cnt1 = cnt0 = 0;
      vector<int> pl, pr;
      for (auto p : t)
        if (p >> i & 1) {
          sum1 += cnt0;
          ++cnt1;
          pl.push_back(p);
        } else {
          sum0 += cnt1;
          ++cnt0;
          pr.push_back(p);
        }
      if (pl.size()) tmp.push_back(pl);
      if (pr.size()) tmp.push_back(pr);
    }
    ans += min(sum0, sum1);
    x |= (sum1 < sum0) << i;
    cur = tmp;
  }
  cout << ans << ' ' << x << endl;
  return 0;
}

```