#include <bits/stdc++.h>
using namespace std;
int amount;
struct trie {
  int child[2];
  vector<int> pass;
};
trie trie_tree[15 * 300005];
int node = 0;
int add_node() {
  node++;
  trie_tree[node].child[0] = trie_tree[node].child[1] = 0;
  trie_tree[node].pass.clear();
  return node;
}
void add_num(int num, int pos) {
  int root = 1;
  for (int bit = 30; bit >= 0; bit--) {
    int now = (num >> bit) & 1;
    if (!trie_tree[root].child[now]) trie_tree[root].child[now] = add_node();
    root = trie_tree[root].child[now];
    trie_tree[root].pass.push_back(pos);
  }
}
long long cal(vector<int> one, vector<int> zero) {
  int zero_size = zero.size(), one_size = one.size();
  long long total = 0;
  int one_st = 0;
  for (int i = 0; i < zero_size; i++) {
    while (one_st < one_size && one[one_st] < zero[i]) one_st++;
    total += one_st;
  }
  return total;
}
long long inverts[35][2];
void process(int root, int bit) {
  if (trie_tree[root].child[0] && trie_tree[root].child[1]) {
    int child_one = trie_tree[root].child[1];
    int child_zero = trie_tree[root].child[0];
    inverts[bit][0] +=
        cal(trie_tree[child_one].pass, trie_tree[child_zero].pass);
    inverts[bit][1] +=
        cal(trie_tree[child_zero].pass, trie_tree[child_one].pass);
  }
  if (trie_tree[root].child[0]) process(trie_tree[root].child[0], bit - 1);
  if (trie_tree[root].child[1]) process(trie_tree[root].child[1], bit - 1);
}
void solve() {
  process(1, 30);
  long long total_inverts = 0;
  int num = 0;
  for (int bit = 0; bit <= 30; bit++) {
    total_inverts += min(inverts[bit][0], inverts[bit][1]);
    if (inverts[bit][0] > inverts[bit][1]) num += 1 << bit;
  }
  printf("%lld %d", total_inverts, num);
}
int main() {
  scanf("%d", &amount);
  add_node();
  for (int i = 1; i <= amount; i++) {
    int numbers;
    scanf("%d", &numbers);
    add_num(numbers, i);
  }
  solve();
}

### Prompt

Your challenge is to write a Cpp solution 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 amount;
struct trie {
  int child[2];
  vector<int> pass;
};
trie trie_tree[15 * 300005];
int node = 0;
int add_node() {
  node++;
  trie_tree[node].child[0] = trie_tree[node].child[1] = 0;
  trie_tree[node].pass.clear();
  return node;
}
void add_num(int num, int pos) {
  int root = 1;
  for (int bit = 30; bit >= 0; bit--) {
    int now = (num >> bit) & 1;
    if (!trie_tree[root].child[now]) trie_tree[root].child[now] = add_node();
    root = trie_tree[root].child[now];
    trie_tree[root].pass.push_back(pos);
  }
}
long long cal(vector<int> one, vector<int> zero) {
  int zero_size = zero.size(), one_size = one.size();
  long long total = 0;
  int one_st = 0;
  for (int i = 0; i < zero_size; i++) {
    while (one_st < one_size && one[one_st] < zero[i]) one_st++;
    total += one_st;
  }
  return total;
}
long long inverts[35][2];
void process(int root, int bit) {
  if (trie_tree[root].child[0] && trie_tree[root].child[1]) {
    int child_one = trie_tree[root].child[1];
    int child_zero = trie_tree[root].child[0];
    inverts[bit][0] +=
        cal(trie_tree[child_one].pass, trie_tree[child_zero].pass);
    inverts[bit][1] +=
        cal(trie_tree[child_zero].pass, trie_tree[child_one].pass);
  }
  if (trie_tree[root].child[0]) process(trie_tree[root].child[0], bit - 1);
  if (trie_tree[root].child[1]) process(trie_tree[root].child[1], bit - 1);
}
void solve() {
  process(1, 30);
  long long total_inverts = 0;
  int num = 0;
  for (int bit = 0; bit <= 30; bit++) {
    total_inverts += min(inverts[bit][0], inverts[bit][1]);
    if (inverts[bit][0] > inverts[bit][1]) num += 1 << bit;
  }
  printf("%lld %d", total_inverts, num);
}
int main() {
  scanf("%d", &amount);
  add_node();
  for (int i = 1; i <= amount; i++) {
    int numbers;
    scanf("%d", &numbers);
    add_num(numbers, i);
  }
  solve();
}

```