#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int64_t total = 0;
  int x = 0;
  vector<vector<int>> vcur = {a};
  for (int j = 29; j >= 0; j--) {
    int64_t sum0 = 0, sum1 = 0;
    vector<vector<int>> vnext;
    for (auto& v : vcur) {
      int cnt0 = 0, cnt1 = 0;
      vector<int> v0, v1;
      for (int b : v) {
        if (b & (1 << j))
          cnt1++, sum1 += cnt0, v1.push_back(b);
        else
          cnt0++, sum0 += cnt1, v0.push_back(b);
      }
      if (v0.size() > 1) vnext.push_back(v0);
      if (v1.size() > 1) vnext.push_back(v1);
    }
    if (sum0 <= sum1)
      total += sum0;
    else
      total += sum1, x |= (1 << j);
    vcur = vnext;
  }
  cout << total << " " << x << endl;
}

### Prompt

In CPP, your task is to solve 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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int64_t total = 0;
  int x = 0;
  vector<vector<int>> vcur = {a};
  for (int j = 29; j >= 0; j--) {
    int64_t sum0 = 0, sum1 = 0;
    vector<vector<int>> vnext;
    for (auto& v : vcur) {
      int cnt0 = 0, cnt1 = 0;
      vector<int> v0, v1;
      for (int b : v) {
        if (b & (1 << j))
          cnt1++, sum1 += cnt0, v1.push_back(b);
        else
          cnt0++, sum0 += cnt1, v0.push_back(b);
      }
      if (v0.size() > 1) vnext.push_back(v0);
      if (v1.size() > 1) vnext.push_back(v1);
    }
    if (sum0 <= sum1)
      total += sum0;
    else
      total += sum1, x |= (1 << j);
    vcur = vnext;
  }
  cout << total << " " << x << endl;
}

```