#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<long long> cur_inv(30, 0);
  vector<long long> fut_inv(30, 0);
  function<void(vector<long long>, long long)> go = [&](vector<long long> xx,
                                                        long long idx) {
    if (xx.empty() || idx < 0) return;
    long long o = 0;
    long long z = 0;
    vector<long long> x1, x2;
    for (auto i : xx) {
      if ((i >> idx) & 1) {
        x1.push_back(i);
        o++;
        fut_inv[idx] += z;
      } else {
        x2.push_back(i);
        z++;
        cur_inv[idx] += o;
      }
    }
    go(x1, idx - 1);
    go(x2, idx - 1);
  };
  go(a, 29);
  long long res = 0;
  long long x = 0;
  for (long long bit = 29; bit >= 0; bit--) {
    if (cur_inv[bit] > fut_inv[bit]) {
      x |= (1 << bit);
    }
    res += min(cur_inv[bit], fut_inv[bit]);
  }
  cout << res << " " << x << "\n";
}

### Prompt

Please create a solution in Cpp to 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;
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<long long> cur_inv(30, 0);
  vector<long long> fut_inv(30, 0);
  function<void(vector<long long>, long long)> go = [&](vector<long long> xx,
                                                        long long idx) {
    if (xx.empty() || idx < 0) return;
    long long o = 0;
    long long z = 0;
    vector<long long> x1, x2;
    for (auto i : xx) {
      if ((i >> idx) & 1) {
        x1.push_back(i);
        o++;
        fut_inv[idx] += z;
      } else {
        x2.push_back(i);
        z++;
        cur_inv[idx] += o;
      }
    }
    go(x1, idx - 1);
    go(x2, idx - 1);
  };
  go(a, 29);
  long long res = 0;
  long long x = 0;
  for (long long bit = 29; bit >= 0; bit--) {
    if (cur_inv[bit] > fut_inv[bit]) {
      x |= (1 << bit);
    }
    res += min(cur_inv[bit], fut_inv[bit]);
  }
  cout << res << " " << x << "\n";
}

```