#include <bits/stdc++.h>
using namespace std;
long long n, a[300001], res, dp[60][3];
void input() {
  freopen(
      ""
      ".inp",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
void Boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
struct Node {
  vector<long long> index;
  int child[3];
};
vector<Node> tree;
void new_node() {
  Node node;
  node.child[0] = -1;
  node.child[1] = -1;
  tree.push_back(node);
}
void add(long long num, long long idx) {
  long long node = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = ((num >> i) & 1);
    if (tree[node].child[bit] == -1) {
      tree[node].child[bit] = tree.size();
      new_node();
    }
    node = tree[node].child[bit];
    tree[node].index.push_back(idx);
  }
}
void get(long long idx, long long node) {
  long long x = tree[node].child[0];
  long long y = tree[node].child[1];
  if (x != -1) get(idx - 1, x);
  if (y != -1) get(idx - 1, y);
  if (x == -1 || y == -1) return;
  long long j = 0, cnt = 0;
  for (auto i : tree[x].index) {
    while (j < (long long)tree[y].index.size() && i > tree[y].index[j]) j++;
    cnt += j;
  }
  dp[idx][0] += cnt;
  dp[idx][1] +=
      ((long long)tree[x].index.size() * (long long)tree[y].index.size()) - cnt;
}
int main() {
  cin >> n;
  new_node();
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    add(a[i], i);
  }
  get(30, 0);
  long long cnt = 0;
  for (int i = 0; i <= 30; i++) {
    cnt += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) {
      res += (1 << i);
    }
  }
  cout << cnt << " " << res;
  return 0;
}

### 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;
long long n, a[300001], res, dp[60][3];
void input() {
  freopen(
      ""
      ".inp",
      "r", stdin);
  freopen(
      ""
      ".out",
      "w", stdout);
}
void Boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
struct Node {
  vector<long long> index;
  int child[3];
};
vector<Node> tree;
void new_node() {
  Node node;
  node.child[0] = -1;
  node.child[1] = -1;
  tree.push_back(node);
}
void add(long long num, long long idx) {
  long long node = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = ((num >> i) & 1);
    if (tree[node].child[bit] == -1) {
      tree[node].child[bit] = tree.size();
      new_node();
    }
    node = tree[node].child[bit];
    tree[node].index.push_back(idx);
  }
}
void get(long long idx, long long node) {
  long long x = tree[node].child[0];
  long long y = tree[node].child[1];
  if (x != -1) get(idx - 1, x);
  if (y != -1) get(idx - 1, y);
  if (x == -1 || y == -1) return;
  long long j = 0, cnt = 0;
  for (auto i : tree[x].index) {
    while (j < (long long)tree[y].index.size() && i > tree[y].index[j]) j++;
    cnt += j;
  }
  dp[idx][0] += cnt;
  dp[idx][1] +=
      ((long long)tree[x].index.size() * (long long)tree[y].index.size()) - cnt;
}
int main() {
  cin >> n;
  new_node();
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    add(a[i], i);
  }
  get(30, 0);
  long long cnt = 0;
  for (int i = 0; i <= 30; i++) {
    cnt += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) {
      res += (1 << i);
    }
  }
  cout << cnt << " " << res;
  return 0;
}

```