#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int> > mass;
  vector<vector<int> > mass1;
  mass1.push_back({});
  for (int i = 0; i < n; i++) {
    mass1[mass1.size() - 1].push_back(i);
  }
  long long allsum = 0;
  int minx = 0;
  for (int i = 29; i >= 0; i--) {
    long long sum1 = 0, sum2 = 0;
    mass.clear();
    for (int j = 0; j < mass1.size(); j++) {
      int cnt0 = 0;
      int cnt1 = 0;
      vector<int> vec1, vec2;
      for (int k = 0; k < mass1[j].size(); k++) {
        if ((a[mass1[j][k]] & (1 << i)) == 0) {
          vec1.push_back(mass1[j][k]);
          sum1 += cnt1;
          cnt0++;
        } else {
          vec2.push_back(mass1[j][k]);
          sum2 += cnt0;
          cnt1++;
        }
      }
      if (vec1.size() != 0) {
        mass.push_back(vec1);
      }
      if (vec2.size() != 0) {
        mass.push_back(vec2);
      }
    }
    allsum += min(sum1, sum2);
    if (sum1 > sum2) {
      minx |= (1 << i);
    }
    mass1.clear();
    for (int j = 0; j < mass.size(); j++) {
      mass1.push_back(mass[j]);
    }
  }
  cout << allsum << " " << minx << "\n";
  return 0;
}

### 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<int> > mass;
  vector<vector<int> > mass1;
  mass1.push_back({});
  for (int i = 0; i < n; i++) {
    mass1[mass1.size() - 1].push_back(i);
  }
  long long allsum = 0;
  int minx = 0;
  for (int i = 29; i >= 0; i--) {
    long long sum1 = 0, sum2 = 0;
    mass.clear();
    for (int j = 0; j < mass1.size(); j++) {
      int cnt0 = 0;
      int cnt1 = 0;
      vector<int> vec1, vec2;
      for (int k = 0; k < mass1[j].size(); k++) {
        if ((a[mass1[j][k]] & (1 << i)) == 0) {
          vec1.push_back(mass1[j][k]);
          sum1 += cnt1;
          cnt0++;
        } else {
          vec2.push_back(mass1[j][k]);
          sum2 += cnt0;
          cnt1++;
        }
      }
      if (vec1.size() != 0) {
        mass.push_back(vec1);
      }
      if (vec2.size() != 0) {
        mass.push_back(vec2);
      }
    }
    allsum += min(sum1, sum2);
    if (sum1 > sum2) {
      minx |= (1 << i);
    }
    mass1.clear();
    for (int j = 0; j < mass.size(); j++) {
      mass1.push_back(mass[j]);
    }
  }
  cout << allsum << " " << minx << "\n";
  return 0;
}

```