#include <bits/stdc++.h>
using namespace std;
long long a[32][2];
void explore(vector<int> v, int b) {
  if (b < 0) return;
  int c0 = 0, c1 = 0;
  long long sum0 = 0, sum1 = 0;
  vector<int> v0, v1;
  for (int i = v.size() - 1; i >= 0; i--) {
    if ((v[i] & (1LL << b))) {
      c1++;
      sum0 += c0;
      v1.push_back(v[i]);
    } else {
      c0++;
      sum1 += c1;
      v0.push_back(v[i]);
    }
  }
  a[b][0] += sum0;
  a[b][1] += sum1;
  reverse(v0.begin(), v0.end());
  reverse(v1.begin(), v1.end());
  if (v0.size() > 1) explore(v0, b - 1);
  if (v1.size() > 1) explore(v1, b - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  explore(v, 30);
  long long res = 0, sum = 0;
  for (int i = 0; i <= 30; i++) {
    sum += min(a[i][0], a[i][1]);
    if (a[i][1] < a[i][0]) (res |= (1LL << i));
  }
  cout << sum << " " << res << "\n";
  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;
long long a[32][2];
void explore(vector<int> v, int b) {
  if (b < 0) return;
  int c0 = 0, c1 = 0;
  long long sum0 = 0, sum1 = 0;
  vector<int> v0, v1;
  for (int i = v.size() - 1; i >= 0; i--) {
    if ((v[i] & (1LL << b))) {
      c1++;
      sum0 += c0;
      v1.push_back(v[i]);
    } else {
      c0++;
      sum1 += c1;
      v0.push_back(v[i]);
    }
  }
  a[b][0] += sum0;
  a[b][1] += sum1;
  reverse(v0.begin(), v0.end());
  reverse(v1.begin(), v1.end());
  if (v0.size() > 1) explore(v0, b - 1);
  if (v1.size() > 1) explore(v1, b - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  explore(v, 30);
  long long res = 0, sum = 0;
  for (int i = 0; i <= 30; i++) {
    sum += min(a[i][0], a[i][1]);
    if (a[i][1] < a[i][0]) (res |= (1LL << i));
  }
  cout << sum << " " << res << "\n";
  return 0;
}

```