#include <bits/stdc++.h>
using namespace std;
int t, n;
int main() {
  vector<deque<int> > lists[2];
  scanf("%d", &n);
  lists[0].push_back({});
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    lists[0][0].push_back(tmp);
  }
  long long total = 0;
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    int mask = 1;
    long long penaltyof0 = 0, penaltyof1 = 0;
    if (i > 0) mask <<= i;
    for (auto that_list : lists[i % 2]) {
      int cnt0 = 0, cnt1 = 0;
      deque<int> zeros, ones;
      for (auto riter = that_list.rbegin(); riter != that_list.rend();
           riter++) {
        if ((*riter) & mask) {
          cnt1++;
          penaltyof0 += cnt0;
          ones.push_front(*riter);
        } else {
          cnt0++;
          penaltyof1 += cnt1;
          zeros.push_front(*riter);
        }
      }
      if (i != 0) {
        if (zeros.size() > 1) lists[(i + 1) % 2].push_back(zeros);
        if (ones.size() > 1) lists[(i + 1) % 2].push_back(ones);
      }
    }
    if (penaltyof0 <= penaltyof1) {
      total += penaltyof0;
    } else {
      total += penaltyof1;
      ans |= mask;
    }
    lists[i % 2].clear();
  }
  printf("%lld %d", total, ans);
}

### Prompt

Your task is to create 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;
int t, n;
int main() {
  vector<deque<int> > lists[2];
  scanf("%d", &n);
  lists[0].push_back({});
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    lists[0][0].push_back(tmp);
  }
  long long total = 0;
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    int mask = 1;
    long long penaltyof0 = 0, penaltyof1 = 0;
    if (i > 0) mask <<= i;
    for (auto that_list : lists[i % 2]) {
      int cnt0 = 0, cnt1 = 0;
      deque<int> zeros, ones;
      for (auto riter = that_list.rbegin(); riter != that_list.rend();
           riter++) {
        if ((*riter) & mask) {
          cnt1++;
          penaltyof0 += cnt0;
          ones.push_front(*riter);
        } else {
          cnt0++;
          penaltyof1 += cnt1;
          zeros.push_front(*riter);
        }
      }
      if (i != 0) {
        if (zeros.size() > 1) lists[(i + 1) % 2].push_back(zeros);
        if (ones.size() > 1) lists[(i + 1) % 2].push_back(ones);
      }
    }
    if (penaltyof0 <= penaltyof1) {
      total += penaltyof0;
    } else {
      total += penaltyof1;
      ans |= mask;
    }
    lists[i % 2].clear();
  }
  printf("%lld %d", total, ans);
}

```