#include <bits/stdc++.h>
using namespace std;
long long ans = 0;
pair<long long, long long> wtf[30];
void recurr(int h, vector<int> V) {
  vector<int> V1, V2;
  int cnt = 0;
  for (int i = 0; i < V.size(); ++i) {
    if ((V[i] >> h) & 1) {
      wtf[h].first += i - cnt;
      ++cnt;
      V1.push_back(V[i]);
    } else
      V2.push_back(V[i]);
  }
  cnt = 0;
  for (int i = V.size() - 1; i >= 0; --i) {
    if ((V[i] >> h) & 1) {
      wtf[h].second += (V.size() - 1 - i) - cnt;
      ++cnt;
    }
  }
  if (h > 0) {
    if (!V1.empty()) recurr(h - 1, V1);
    if (!V2.empty()) recurr(h - 1, V2);
  }
}
int main() {
  long long n, x = 0;
  cin >> n;
  vector<int> V;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    V.push_back(a);
  }
  for (int i = 0; i < 30; ++i) {
    wtf[i] = {0, 0};
  }
  recurr(29, V);
  for (int i = 0; i < 30; ++i) {
    ans += min(wtf[i].first, wtf[i].second);
    if (wtf[i].first < wtf[i].second) {
      x += (1 << i);
    }
  }
  cout << ans << ' ' << x << '\n';
}

### Prompt

Generate a Cpp solution 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;
long long ans = 0;
pair<long long, long long> wtf[30];
void recurr(int h, vector<int> V) {
  vector<int> V1, V2;
  int cnt = 0;
  for (int i = 0; i < V.size(); ++i) {
    if ((V[i] >> h) & 1) {
      wtf[h].first += i - cnt;
      ++cnt;
      V1.push_back(V[i]);
    } else
      V2.push_back(V[i]);
  }
  cnt = 0;
  for (int i = V.size() - 1; i >= 0; --i) {
    if ((V[i] >> h) & 1) {
      wtf[h].second += (V.size() - 1 - i) - cnt;
      ++cnt;
    }
  }
  if (h > 0) {
    if (!V1.empty()) recurr(h - 1, V1);
    if (!V2.empty()) recurr(h - 1, V2);
  }
}
int main() {
  long long n, x = 0;
  cin >> n;
  vector<int> V;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    V.push_back(a);
  }
  for (int i = 0; i < 30; ++i) {
    wtf[i] = {0, 0};
  }
  recurr(29, V);
  for (int i = 0; i < 30; ++i) {
    ans += min(wtf[i].first, wtf[i].second);
    if (wtf[i].first < wtf[i].second) {
      x += (1 << i);
    }
  }
  cout << ans << ' ' << x << '\n';
}

```