#include <bits/stdc++.h>
using namespace std;
int tree[300020];
int n;
void update(int pos, int val) {
  pos++;
  while (pos <= n) {
    tree[pos] += val;
    pos += pos & -pos;
  }
}
int query(int r) {
  r++;
  int res = 0;
  while (r > 0) {
    res += tree[r];
    r -= r & -r;
  }
  return res;
}
long long getInvs(vector<int> arr, int x) {
  memset(tree, 0, sizeof(tree));
  for_each(arr.begin(), arr.end(), [&](int &el) { el ^= x; });
  vector<pair<int, int>> compr(n);
  for (int i = 0; i < n; i++) {
    compr[i] = {arr[i], i};
  }
  sort(compr.begin(), compr.end());
  int c = 1;
  for (int i = 0; i < n; i++) {
    if (i > 0 && compr[i - 1].first < compr[i].first) c++;
    arr[compr[i].second] = c;
  }
  long long answer = 0;
  for (int i = n - 1; i >= 0; i--) {
    answer += query(arr[i] - 1);
    update(arr[i], 1);
  }
  return answer;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<int> arr(n);
  for (int &it : arr) {
    cin >> it;
  }
  int x = 0;
  long long invs = getInvs(arr, 0);
  for (int i = 1 << 30; i > 0; i >>= 1) {
    x ^= i;
    long long newInvs = getInvs(arr, x);
    if (newInvs < invs) {
      invs = newInvs;
    } else {
      x ^= i;
    }
  }
  cout << invs << " " << x << '\n';
}

### 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 tree[300020];
int n;
void update(int pos, int val) {
  pos++;
  while (pos <= n) {
    tree[pos] += val;
    pos += pos & -pos;
  }
}
int query(int r) {
  r++;
  int res = 0;
  while (r > 0) {
    res += tree[r];
    r -= r & -r;
  }
  return res;
}
long long getInvs(vector<int> arr, int x) {
  memset(tree, 0, sizeof(tree));
  for_each(arr.begin(), arr.end(), [&](int &el) { el ^= x; });
  vector<pair<int, int>> compr(n);
  for (int i = 0; i < n; i++) {
    compr[i] = {arr[i], i};
  }
  sort(compr.begin(), compr.end());
  int c = 1;
  for (int i = 0; i < n; i++) {
    if (i > 0 && compr[i - 1].first < compr[i].first) c++;
    arr[compr[i].second] = c;
  }
  long long answer = 0;
  for (int i = n - 1; i >= 0; i--) {
    answer += query(arr[i] - 1);
    update(arr[i], 1);
  }
  return answer;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<int> arr(n);
  for (int &it : arr) {
    cin >> it;
  }
  int x = 0;
  long long invs = getInvs(arr, 0);
  for (int i = 1 << 30; i > 0; i >>= 1) {
    x ^= i;
    long long newInvs = getInvs(arr, x);
    if (newInvs < invs) {
      invs = newInvs;
    } else {
      x ^= i;
    }
  }
  cout << invs << " " << x << '\n';
}

```