#include <bits/stdc++.h>
using namespace std;
long long t = 1;
template <class myType>
void print_arr(myType &arr, long long L, string sep) {
  for (long long(i) = (0); (i) < (L); ++(i)) {
    cout << arr[i];
    if (i < L - 1) {
      cout << sep;
    }
  }
  cout << "\n";
  return;
}
long long merge(long long arr[], long long temp_arr[], long long left,
                long long mid, long long right) {
  long long i, j, k;
  i = left;
  j = mid + 1;
  k = left;
  long long inv_count = 0;
  while (i <= mid && j <= right) {
    if (arr[i] <= arr[j]) {
      temp_arr[k] = arr[i];
      k += 1;
      i += 1;
    } else {
      temp_arr[k] = arr[j];
      inv_count += (mid - i + 1);
      k += 1;
      j += 1;
    }
  }
  while (i <= mid) {
    temp_arr[k] = arr[i];
    k += 1;
    i += 1;
  }
  while (j <= right) {
    temp_arr[k] = arr[j];
    k += 1;
    j += 1;
  }
  for (long long loop_var = left; loop_var <= right; loop_var++) {
    arr[loop_var] = temp_arr[loop_var];
  }
  return inv_count;
}
long long _mergeSort(long long arr[], long long temp_arr[], long long left,
                     long long right) {
  long long inv_count = 0;
  if (left < right) {
    long long mid;
    mid = (left + right) / 2;
    inv_count += _mergeSort(arr, temp_arr, left, mid);
    inv_count += _mergeSort(arr, temp_arr, mid + 1, right);
    inv_count += merge(arr, temp_arr, left, mid, right);
  }
  return inv_count;
}
long long mergeSort(long long arr[], long long n) {
  long long temp_arr[n];
  for (long long(i) = (0); (i) < (n); ++(i)) temp_arr[i] = 0;
  return _mergeSort(arr, temp_arr, 0, n - 1);
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long(i) = (0); (i) < (n); ++(i)) cin >> a[i];
  long long invs, invs2;
  long long b[n];
  for (long long(i) = (0); (i) < (n); ++(i)) b[i] = a[i];
  invs = mergeSort(b, n);
  long long ans = 0;
  long long p;
  for (long long(i) = (0); (i) < (30); ++(i)) {
    p = (1 << i);
    for (long long(j) = (0); (j) < (n); ++(j)) b[j] = a[j] ^ p;
    invs2 = mergeSort(b, n);
    if (invs2 < invs) {
      ans += p;
    }
  }
  for (long long(j) = (0); (j) < (n); ++(j)) b[j] = a[j] ^ ans;
  cout << mergeSort(b, n) << " " << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 t = 1;
template <class myType>
void print_arr(myType &arr, long long L, string sep) {
  for (long long(i) = (0); (i) < (L); ++(i)) {
    cout << arr[i];
    if (i < L - 1) {
      cout << sep;
    }
  }
  cout << "\n";
  return;
}
long long merge(long long arr[], long long temp_arr[], long long left,
                long long mid, long long right) {
  long long i, j, k;
  i = left;
  j = mid + 1;
  k = left;
  long long inv_count = 0;
  while (i <= mid && j <= right) {
    if (arr[i] <= arr[j]) {
      temp_arr[k] = arr[i];
      k += 1;
      i += 1;
    } else {
      temp_arr[k] = arr[j];
      inv_count += (mid - i + 1);
      k += 1;
      j += 1;
    }
  }
  while (i <= mid) {
    temp_arr[k] = arr[i];
    k += 1;
    i += 1;
  }
  while (j <= right) {
    temp_arr[k] = arr[j];
    k += 1;
    j += 1;
  }
  for (long long loop_var = left; loop_var <= right; loop_var++) {
    arr[loop_var] = temp_arr[loop_var];
  }
  return inv_count;
}
long long _mergeSort(long long arr[], long long temp_arr[], long long left,
                     long long right) {
  long long inv_count = 0;
  if (left < right) {
    long long mid;
    mid = (left + right) / 2;
    inv_count += _mergeSort(arr, temp_arr, left, mid);
    inv_count += _mergeSort(arr, temp_arr, mid + 1, right);
    inv_count += merge(arr, temp_arr, left, mid, right);
  }
  return inv_count;
}
long long mergeSort(long long arr[], long long n) {
  long long temp_arr[n];
  for (long long(i) = (0); (i) < (n); ++(i)) temp_arr[i] = 0;
  return _mergeSort(arr, temp_arr, 0, n - 1);
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long(i) = (0); (i) < (n); ++(i)) cin >> a[i];
  long long invs, invs2;
  long long b[n];
  for (long long(i) = (0); (i) < (n); ++(i)) b[i] = a[i];
  invs = mergeSort(b, n);
  long long ans = 0;
  long long p;
  for (long long(i) = (0); (i) < (30); ++(i)) {
    p = (1 << i);
    for (long long(j) = (0); (j) < (n); ++(j)) b[j] = a[j] ^ p;
    invs2 = mergeSort(b, n);
    if (invs2 < invs) {
      ans += p;
    }
  }
  for (long long(j) = (0); (j) < (n); ++(j)) b[j] = a[j] ^ ans;
  cout << mergeSort(b, n) << " " << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  while (t--) {
    solve();
  }
  return 0;
}

```