#include <bits/stdc++.h>
using namespace std;
int temp[300005];
long long _mergeSort(int arr[], int temp[], int left, int right);
long long merge(int arr[], int temp[], int left, int mid, int right);
long long mergeSort(int arr[], int array_size) {
  return _mergeSort(arr, temp, 0, array_size - 1);
}
long long _mergeSort(int arr[], 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 merge(int arr[], 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;
}
int a[300005];
int b[300005];
long long invCount(int a[], int n) {
  for (int i = 0; i < n; ++i) {
    b[i] = a[i];
  }
  return mergeSort(b, n);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  bool fresh = true;
  long long ic0 = invCount(a, n);
  long long ans = 0;
  for (int bit = 29; bit >= 0; --bit) {
    if (!fresh) {
      ic0 = invCount(a, n);
      fresh = true;
    }
    for (int i = 0; i < n; ++i) {
      a[i] ^= (1 << bit);
    }
    long long ic1 = invCount(a, n);
    if (ic1 < ic0) {
      ans ^= (1 << bit);
      fresh = false;
      continue;
    }
    for (int i = 0; i < n; ++i) {
      a[i] ^= (1 << bit);
    }
  }
  cout << invCount(a, n) << " " << ans;
}

### 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;
int temp[300005];
long long _mergeSort(int arr[], int temp[], int left, int right);
long long merge(int arr[], int temp[], int left, int mid, int right);
long long mergeSort(int arr[], int array_size) {
  return _mergeSort(arr, temp, 0, array_size - 1);
}
long long _mergeSort(int arr[], 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 merge(int arr[], 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;
}
int a[300005];
int b[300005];
long long invCount(int a[], int n) {
  for (int i = 0; i < n; ++i) {
    b[i] = a[i];
  }
  return mergeSort(b, n);
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  bool fresh = true;
  long long ic0 = invCount(a, n);
  long long ans = 0;
  for (int bit = 29; bit >= 0; --bit) {
    if (!fresh) {
      ic0 = invCount(a, n);
      fresh = true;
    }
    for (int i = 0; i < n; ++i) {
      a[i] ^= (1 << bit);
    }
    long long ic1 = invCount(a, n);
    if (ic1 < ic0) {
      ans ^= (1 << bit);
      fresh = false;
      continue;
    }
    for (int i = 0; i < n; ++i) {
      a[i] ^= (1 << bit);
    }
  }
  cout << invCount(a, n) << " " << ans;
}

```