#include <bits/stdc++.h>
using namespace std;
long long merge(vector<int>& arr, vector<int>& temp, int left, int mid,
                int right) {
  int i, j, k;
  long long inv_count = 0;
  i = left;
  j = mid;
  k = left;
  while ((i <= mid - 1) && (j <= right)) {
    if (arr[i] <= arr[j]) {
      temp[k++] = arr[i++];
    } else {
      temp[k++] = arr[j++];
      inv_count = inv_count + (mid - i);
    }
  }
  while (i <= mid - 1) temp[k++] = arr[i++];
  while (j <= right) temp[k++] = arr[j++];
  for (i = left; i <= right; i++) arr[i] = temp[i];
  return inv_count;
}
long long _mergeSort(vector<int>& arr, vector<int>& temp, int left, int right) {
  int mid;
  long long inv_count = 0;
  if (right > left) {
    mid = (right + left) / 2;
    inv_count += _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid + 1, right);
    inv_count += merge(arr, temp, left, mid + 1, right);
  }
  return inv_count;
}
long long count_inv(vector<int> v) {
  vector<int> temp(v.size());
  return _mergeSort(v, temp, 0, v.size() - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < (int)n; i++) cin >> v[i];
  long long cur = count_inv(v);
  int ans = 0;
  for (int b = 29; b >= 0; b--) {
    int st = (1 << b);
    for (int i = 0; i < (int)n; i++) v[i] ^= st;
    long long nw = count_inv(v);
    if (nw < cur) {
      cur = nw;
      ans |= st;
    } else {
      for (int i = 0; i < (int)n; i++) v[i] ^= st;
    }
  }
  cout << cur << " " << ans << "\n";
}

### Prompt

Please formulate 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 merge(vector<int>& arr, vector<int>& temp, int left, int mid,
                int right) {
  int i, j, k;
  long long inv_count = 0;
  i = left;
  j = mid;
  k = left;
  while ((i <= mid - 1) && (j <= right)) {
    if (arr[i] <= arr[j]) {
      temp[k++] = arr[i++];
    } else {
      temp[k++] = arr[j++];
      inv_count = inv_count + (mid - i);
    }
  }
  while (i <= mid - 1) temp[k++] = arr[i++];
  while (j <= right) temp[k++] = arr[j++];
  for (i = left; i <= right; i++) arr[i] = temp[i];
  return inv_count;
}
long long _mergeSort(vector<int>& arr, vector<int>& temp, int left, int right) {
  int mid;
  long long inv_count = 0;
  if (right > left) {
    mid = (right + left) / 2;
    inv_count += _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid + 1, right);
    inv_count += merge(arr, temp, left, mid + 1, right);
  }
  return inv_count;
}
long long count_inv(vector<int> v) {
  vector<int> temp(v.size());
  return _mergeSort(v, temp, 0, v.size() - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < (int)n; i++) cin >> v[i];
  long long cur = count_inv(v);
  int ans = 0;
  for (int b = 29; b >= 0; b--) {
    int st = (1 << b);
    for (int i = 0; i < (int)n; i++) v[i] ^= st;
    long long nw = count_inv(v);
    if (nw < cur) {
      cur = nw;
      ans |= st;
    } else {
      for (int i = 0; i < (int)n; i++) v[i] ^= st;
    }
  }
  cout << cur << " " << ans << "\n";
}

```