#include <bits/stdc++.h>
using namespace std;
bool check(vector<long long int> &a, long long int k, long long int n) {
  if (a.front() > k) {
    return false;
  }
  if (a.back() <= n - k) {
    return false;
  }
  for (long long int i = 1; i < a.size(); i++) {
    if (a[i] - a[i - 1] > k) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<bitset<31>>> vec(1), temp;
  for (long long int i = 0; i < n; i++) {
    bitset<31> T;
    T = a[i];
    vec[0].push_back(T);
  }
  bitset<31> X;
  long long int ans = 0;
  for (long long int bit = 30; bit >= 0; bit--) {
    map<int, int> mp;
    long long int a1 = 0, a2 = 0;
    temp.clear();
    for (long long int i = 0; i < vec.size(); i++) {
      mp.clear();
      for (long long int j = 0; j < vec[i].size(); j++) {
        if (vec[i][j][bit]) {
          a1 += mp[0];
          mp[1]++;
        } else {
          a2 += mp[1];
          mp[0]++;
        }
      }
    }
    if (a1 < a2) {
      X[bit] = true;
      ans += a1;
    } else {
      X[bit] = false;
      ans += a2;
    }
    for (long long int i = 0; i < vec.size(); i++) {
      vector<bitset<31>> TT1, TT2;
      for (long long int j = 0; j < vec[i].size(); j++) {
        if (vec[i][j][bit]) {
          TT1.push_back(vec[i][j]);
        } else {
          TT2.push_back(vec[i][j]);
        }
      }
      if (TT1.size()) temp.push_back(TT1);
      if (TT2.size()) temp.push_back(TT2);
    }
    vec = temp;
  }
  cout << ans << " " << X.to_ullong() << "\n";
}

### Prompt

Please create a solution in cpp 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;
bool check(vector<long long int> &a, long long int k, long long int n) {
  if (a.front() > k) {
    return false;
  }
  if (a.back() <= n - k) {
    return false;
  }
  for (long long int i = 1; i < a.size(); i++) {
    if (a[i] - a[i - 1] > k) {
      return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<bitset<31>>> vec(1), temp;
  for (long long int i = 0; i < n; i++) {
    bitset<31> T;
    T = a[i];
    vec[0].push_back(T);
  }
  bitset<31> X;
  long long int ans = 0;
  for (long long int bit = 30; bit >= 0; bit--) {
    map<int, int> mp;
    long long int a1 = 0, a2 = 0;
    temp.clear();
    for (long long int i = 0; i < vec.size(); i++) {
      mp.clear();
      for (long long int j = 0; j < vec[i].size(); j++) {
        if (vec[i][j][bit]) {
          a1 += mp[0];
          mp[1]++;
        } else {
          a2 += mp[1];
          mp[0]++;
        }
      }
    }
    if (a1 < a2) {
      X[bit] = true;
      ans += a1;
    } else {
      X[bit] = false;
      ans += a2;
    }
    for (long long int i = 0; i < vec.size(); i++) {
      vector<bitset<31>> TT1, TT2;
      for (long long int j = 0; j < vec[i].size(); j++) {
        if (vec[i][j][bit]) {
          TT1.push_back(vec[i][j]);
        } else {
          TT2.push_back(vec[i][j]);
        }
      }
      if (TT1.size()) temp.push_back(TT1);
      if (TT2.size()) temp.push_back(TT2);
    }
    vec = temp;
  }
  cout << ans << " " << X.to_ullong() << "\n";
}

```