#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::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];
  long long ans = 0, ans2 = 0;
  vector<vector<long long>> tmp = {a};
  for (long long j = 30; j >= 0; j--) {
    vector<vector<long long>> tmp2;
    long long c = 0, c2 = 0;
    for (auto a : tmp) {
      vector<long long> X, Y;
      long long v = 0, v2 = 0;
      for (auto x : a) {
        if ((x & (1 << j))) {
          c2 += v;
          v2++;
          X.push_back(x);
        } else {
          c += v2;
          v++;
          Y.push_back(x);
        }
      }
      if (X.size() > 0) tmp2.push_back(X);
      if (Y.size() > 0) tmp2.push_back(Y);
    }
    ans += min(c, c2);
    if (c2 < c) ans2 += (1 << j);
    tmp = tmp2;
  }
  cout << ans << " " << ans2;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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_base::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];
  long long ans = 0, ans2 = 0;
  vector<vector<long long>> tmp = {a};
  for (long long j = 30; j >= 0; j--) {
    vector<vector<long long>> tmp2;
    long long c = 0, c2 = 0;
    for (auto a : tmp) {
      vector<long long> X, Y;
      long long v = 0, v2 = 0;
      for (auto x : a) {
        if ((x & (1 << j))) {
          c2 += v;
          v2++;
          X.push_back(x);
        } else {
          c += v2;
          v++;
          Y.push_back(x);
        }
      }
      if (X.size() > 0) tmp2.push_back(X);
      if (Y.size() > 0) tmp2.push_back(Y);
    }
    ans += min(c, c2);
    if (c2 < c) ans2 += (1 << j);
    tmp = tmp2;
  }
  cout << ans << " " << ans2;
}

```