#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  int n;
  cin >> n;
  vector<int> A(n);
  for (int i = 0; i < n; ++i) cin >> A[i];
  vector<int> B;
  for (int i = 0; i < n; ++i) B.push_back(i);
  vector<long long> C(31), D(31);
  function<void(vector<int>, int)> dfs = [&](vector<int> S, int cur) {
    if (cur == -1 || ((int)(S).size()) == 0) return;
    vector<int> S1, S2;
    long long res = 0, o = 0, res2 = 0, o2 = 0;
    for (int i = 0; i < ((int)(S).size()); ++i) {
      if (A[S[i]] >> cur & 1) {
        o++;
        S2.push_back(S[i]);
        res2 += o2;
      } else {
        res += o;
        o2++;
        S1.push_back(S[i]);
      }
    }
    C[cur] += res;
    D[cur] += res2;
    dfs(S1, cur - 1);
    dfs(S2, cur - 1);
  };
  dfs(B, 30);
  int res = 0;
  long long ans = 0;
  for (int i = 30; i >= 0; --i) {
    if (C[i] > D[i]) {
      res |= 1 << i;
      ans += D[i];
    } else {
      ans += C[i];
    }
  }
  cout << ans << " " << res << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  int n;
  cin >> n;
  vector<int> A(n);
  for (int i = 0; i < n; ++i) cin >> A[i];
  vector<int> B;
  for (int i = 0; i < n; ++i) B.push_back(i);
  vector<long long> C(31), D(31);
  function<void(vector<int>, int)> dfs = [&](vector<int> S, int cur) {
    if (cur == -1 || ((int)(S).size()) == 0) return;
    vector<int> S1, S2;
    long long res = 0, o = 0, res2 = 0, o2 = 0;
    for (int i = 0; i < ((int)(S).size()); ++i) {
      if (A[S[i]] >> cur & 1) {
        o++;
        S2.push_back(S[i]);
        res2 += o2;
      } else {
        res += o;
        o2++;
        S1.push_back(S[i]);
      }
    }
    C[cur] += res;
    D[cur] += res2;
    dfs(S1, cur - 1);
    dfs(S2, cur - 1);
  };
  dfs(B, 30);
  int res = 0;
  long long ans = 0;
  for (int i = 30; i >= 0; --i) {
    if (C[i] > D[i]) {
      res |= 1 << i;
      ans += D[i];
    } else {
      ans += C[i];
    }
  }
  cout << ans << " " << res << "\n";
  return 0;
}

```