#include <bits/stdc++.h>
using namespace std;
int get_bit(int val, int id) { return (val >> id) & 1; }
struct node {
  int num;
  int child[2];
};
long long valInv[30 + 5][2];
vector<node> trie;
void new_node() {
  node New;
  New.num = 0;
  for (int i = 0; i <= 1; i++) New.child[i] = -1;
  trie.push_back(New);
}
void add(int val) {
  int indx = 0;
  for (int bit = 30; bit >= 0; bit--) {
    int x = get_bit(val, bit);
    trie[indx].num++;
    if (trie[indx].child[x] == -1) {
      new_node();
      trie[indx].child[x] = trie.size() - 1;
    }
    if (trie[indx].child[1 - x] >= 0)
      valInv[bit][x] += trie[trie[indx].child[1 - x]].num;
    indx = trie[indx].child[x];
  }
  trie[indx].num++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  new_node();
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  long long res = 0, resVal = 0;
  for (int bit = 0; bit <= 30; bit++) {
    res += min(valInv[bit][0], valInv[bit][1]);
    if (valInv[bit][0] > valInv[bit][1]) {
      resVal += (1 << bit);
    }
  }
  cout << res << ' ' << resVal << '\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;
int get_bit(int val, int id) { return (val >> id) & 1; }
struct node {
  int num;
  int child[2];
};
long long valInv[30 + 5][2];
vector<node> trie;
void new_node() {
  node New;
  New.num = 0;
  for (int i = 0; i <= 1; i++) New.child[i] = -1;
  trie.push_back(New);
}
void add(int val) {
  int indx = 0;
  for (int bit = 30; bit >= 0; bit--) {
    int x = get_bit(val, bit);
    trie[indx].num++;
    if (trie[indx].child[x] == -1) {
      new_node();
      trie[indx].child[x] = trie.size() - 1;
    }
    if (trie[indx].child[1 - x] >= 0)
      valInv[bit][x] += trie[trie[indx].child[1 - x]].num;
    indx = trie[indx].child[x];
  }
  trie[indx].num++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  new_node();
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  long long res = 0, resVal = 0;
  for (int bit = 0; bit <= 30; bit++) {
    res += min(valInv[bit][0], valInv[bit][1]);
    if (valInv[bit][0] > valInv[bit][1]) {
      resVal += (1 << bit);
    }
  }
  cout << res << ' ' << resVal << '\n';
}

```