#include <bits/stdc++.h>
using namespace std;
map<pair<int, int>, long long> m;
void rec(vector<int> &a, int i) {
  if (a.size() == 0) {
    m[{i, 0}] += 0;
    m[{i, 1}] += 0;
    return;
  }
  if (i == -1) return;
  vector<int> o, z;
  long long inv = 0;
  int n1 = 0;
  long long temp = (1 << i);
  int n = a.size();
  for (int j = 0; j < n; j++) {
    if (a[j] & temp) {
      o.push_back(a[j]);
      n1++;
    } else {
      z.push_back(a[j]);
      inv += n1;
    }
  }
  m[{i, 0}] += inv;
  inv = 0;
  n1 = 0;
  for (int j = 0; j < n; j++) {
    if (a[j] & temp)
      inv += n1;
    else
      n1++;
  }
  m[{i, 1}] += inv;
  rec(o, i - 1);
  rec(z, i - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  int mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  int x = 0;
  int msb = -1;
  if (mx != 0) msb = log2(mx);
  long long inv = 0;
  rec(a, msb);
  for (int i = 0; i <= msb; i++) {
    if (m[{i, 0}] > m[{i, 1}]) x += pow(2, i);
    inv += min(m[{i, 0}], m[{i, 1}]);
  }
  cout << inv << " " << x << endl;
}

### 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;
map<pair<int, int>, long long> m;
void rec(vector<int> &a, int i) {
  if (a.size() == 0) {
    m[{i, 0}] += 0;
    m[{i, 1}] += 0;
    return;
  }
  if (i == -1) return;
  vector<int> o, z;
  long long inv = 0;
  int n1 = 0;
  long long temp = (1 << i);
  int n = a.size();
  for (int j = 0; j < n; j++) {
    if (a[j] & temp) {
      o.push_back(a[j]);
      n1++;
    } else {
      z.push_back(a[j]);
      inv += n1;
    }
  }
  m[{i, 0}] += inv;
  inv = 0;
  n1 = 0;
  for (int j = 0; j < n; j++) {
    if (a[j] & temp)
      inv += n1;
    else
      n1++;
  }
  m[{i, 1}] += inv;
  rec(o, i - 1);
  rec(z, i - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  int mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  int x = 0;
  int msb = -1;
  if (mx != 0) msb = log2(mx);
  long long inv = 0;
  rec(a, msb);
  for (int i = 0; i <= msb; i++) {
    if (m[{i, 0}] > m[{i, 1}]) x += pow(2, i);
    inv += min(m[{i, 0}], m[{i, 1}]);
  }
  cout << inv << " " << x << endl;
}

```