#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
long long pls[31];
long long mis[31];
void solve(int k, vector<int> &v) {
  if (k == -1) return;
  vector<int> a, b;
  for (int i = 0; i < (v.size()); ++i) {
    if (v[i] & (1 << k)) {
      pls[k] += (int)b.size();
      a.push_back(v[i]);
    } else {
      mis[k] += (int)a.size();
      b.push_back(v[i]);
    }
  }
  if (a.size()) solve(k - 1, a);
  if (b.size()) solve(k - 1, b);
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> A(N);
  for (int i = 0; i < (N); ++i) cin >> A[i];
  vector<int> B = A;
  solve(30, B);
  long long ansb = 0, ansx = 0;
  for (int k = 30; k >= 0; k--) {
    if (pls[k] == 0 && mis[k] == 0) continue;
    if (pls[k] >= mis[k]) {
      ansb += mis[k];
    } else {
      ansb += pls[k];
      ansx += (1 << k);
    }
  }
  cout << ansb << " " << ansx << endl;
  return 0;
}

### 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;
const long long MOD = 1e9 + 7;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
long long pls[31];
long long mis[31];
void solve(int k, vector<int> &v) {
  if (k == -1) return;
  vector<int> a, b;
  for (int i = 0; i < (v.size()); ++i) {
    if (v[i] & (1 << k)) {
      pls[k] += (int)b.size();
      a.push_back(v[i]);
    } else {
      mis[k] += (int)a.size();
      b.push_back(v[i]);
    }
  }
  if (a.size()) solve(k - 1, a);
  if (b.size()) solve(k - 1, b);
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> A(N);
  for (int i = 0; i < (N); ++i) cin >> A[i];
  vector<int> B = A;
  solve(30, B);
  long long ansb = 0, ansx = 0;
  for (int k = 30; k >= 0; k--) {
    if (pls[k] == 0 && mis[k] == 0) continue;
    if (pls[k] >= mis[k]) {
      ansb += mis[k];
    } else {
      ansb += pls[k];
      ansx += (1 << k);
    }
  }
  cout << ansb << " " << ansx << endl;
  return 0;
}

```