#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long double PI = acos((long double)-1);
long long cnt[32][2];
class Node {
 public:
  Node* child[2];
  long long cnt;
  Node() {
    child[0] = child[1] = NULL;
    cnt = 0;
  }
};
class Trie {
 private:
  Node* root;

 public:
  Trie() { root = new Node(); }
  void insert(long long no) {
    Node* curr = root;
    for (long long i = 31; i >= 0; --i) {
      long long bit = (no >> i) & 1;
      if (!curr->child[bit]) {
        curr->child[bit] = new Node();
      }
      if (curr->child[0] && curr->child[1]) {
        if (bit) {
          cnt[i][bit] += curr->child[0]->cnt;
        } else {
          cnt[i][bit] += curr->child[1]->cnt;
        }
      }
      curr = curr->child[bit];
      ++curr->cnt;
    }
  }
};
void solve() {
  long long n;
  cin >> n;
  Trie t;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    t.insert(x);
  }
  long long x = 0;
  long long inv = 0;
  for (long long i = 31; i >= 0; --i) {
    if (cnt[i][1] < cnt[i][0]) {
      x += (1 << i);
      inv += cnt[i][1];
    } else {
      inv += cnt[i][0];
    }
  }
  cout << inv << " " << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 MOD = 1e9 + 7;
const long double PI = acos((long double)-1);
long long cnt[32][2];
class Node {
 public:
  Node* child[2];
  long long cnt;
  Node() {
    child[0] = child[1] = NULL;
    cnt = 0;
  }
};
class Trie {
 private:
  Node* root;

 public:
  Trie() { root = new Node(); }
  void insert(long long no) {
    Node* curr = root;
    for (long long i = 31; i >= 0; --i) {
      long long bit = (no >> i) & 1;
      if (!curr->child[bit]) {
        curr->child[bit] = new Node();
      }
      if (curr->child[0] && curr->child[1]) {
        if (bit) {
          cnt[i][bit] += curr->child[0]->cnt;
        } else {
          cnt[i][bit] += curr->child[1]->cnt;
        }
      }
      curr = curr->child[bit];
      ++curr->cnt;
    }
  }
};
void solve() {
  long long n;
  cin >> n;
  Trie t;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    t.insert(x);
  }
  long long x = 0;
  long long inv = 0;
  for (long long i = 31; i >= 0; --i) {
    if (cnt[i][1] < cnt[i][0]) {
      x += (1 << i);
      inv += cnt[i][1];
    } else {
      inv += cnt[i][0];
    }
  }
  cout << inv << " " << x << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}

```