#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 113;
int val[N];
pair<long long, long long> solve(vector<int> &v, int b) {
  long long c1 = 0, c2 = 0, sum1 = 0, sum2 = 0;
  for (auto i : v) {
    if ((val[i] & b))
      c1++, sum2 += c2;
    else
      c2++, sum1 += c1;
  }
  return {sum1, sum2};
}
signed main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> v(n);
  vector<vector<int>> elem;
  elem.push_back(vector<int>());
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    val[i] = v[i];
    elem.back().push_back(i);
  }
  long long ans = 0, sum = 0;
  for (int i = (1 << 30); i > 0; i /= 2) {
    long long val1 = 0, val2 = 0;
    for (int j = 0; j < elem.size(); j++) {
      auto s = solve(elem[j], i);
      val1 += s.first, val2 += s.second;
    }
    if (val1 > val2) ans += i;
    sum += min(val1, val2);
    vector<vector<int>> newel;
    for (int j = 0; j < elem.size(); j++) {
      vector<int> v1, v2;
      for (auto k : elem[j]) {
        if (val[k] & i)
          v1.push_back(k);
        else
          v2.push_back(k);
      }
      if (v1.size()) newel.push_back(v1);
      if (v2.size()) newel.push_back(v2);
    }
    elem = newel;
  }
  cout << sum << " " << ans << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 3e5 + 113;
int val[N];
pair<long long, long long> solve(vector<int> &v, int b) {
  long long c1 = 0, c2 = 0, sum1 = 0, sum2 = 0;
  for (auto i : v) {
    if ((val[i] & b))
      c1++, sum2 += c2;
    else
      c2++, sum1 += c1;
  }
  return {sum1, sum2};
}
signed main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> v(n);
  vector<vector<int>> elem;
  elem.push_back(vector<int>());
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    val[i] = v[i];
    elem.back().push_back(i);
  }
  long long ans = 0, sum = 0;
  for (int i = (1 << 30); i > 0; i /= 2) {
    long long val1 = 0, val2 = 0;
    for (int j = 0; j < elem.size(); j++) {
      auto s = solve(elem[j], i);
      val1 += s.first, val2 += s.second;
    }
    if (val1 > val2) ans += i;
    sum += min(val1, val2);
    vector<vector<int>> newel;
    for (int j = 0; j < elem.size(); j++) {
      vector<int> v1, v2;
      for (auto k : elem[j]) {
        if (val[k] & i)
          v1.push_back(k);
        else
          v2.push_back(k);
      }
      if (v1.size()) newel.push_back(v1);
      if (v2.size()) newel.push_back(v2);
    }
    elem = newel;
  }
  cout << sum << " " << ans << endl;
}

```