#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const long long maxc = 2e9 + 7;
const int LOG = 30;
int n, a[maxn];
struct Node {
  Node *child[2];
  vector<int> *id;
  Node() {
    memset(child, (0), sizeof(child));
    id = new vector<int>();
  }
};
Node *root = new Node();
void add(int x, int pos) {
  Node *p = root;
  for (int i = (LOG); i >= (0); --i) {
    int k = (((x) >> (i)) & 1);
    if (p->child[k] == 0) p->child[k] = new Node();
    p = p->child[k];
    p->id->push_back(pos);
  }
}
long long dp[LOG + 1][2];
void go(Node *root, int level = LOG) {
  if (level < 0) return;
  if (root->child[0] != NULL) go(root->child[0], level - 1);
  if (root->child[1] != NULL) go(root->child[1], level - 1);
  if (root->child[0] == NULL || root->child[1] == NULL) return;
  long long res = 0;
  for (auto x : *root->child[0]->id) {
    int t = lower_bound((*root->child[1]->id).begin(),
                        (*root->child[1]->id).end(), x) -
            root->child[1]->id->begin() - 1;
    res += 1LL + t;
  }
  dp[level][0] += res;
  dp[level][1] +=
      1LL * (root->child[0]->id->size()) * (root->child[1]->id->size()) - res;
}
void solve() {
  go(root);
  long long ans = 0;
  int res = 0;
  for (int level = (0); level <= (LOG); ++level) {
    ans += 1LL * min(dp[level][0], dp[level][1]);
    if (dp[level][1] < dp[level][0]) res = res ^ (1ll << (level));
  }
  cout << ans << " " << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) add(a[i], i);
  solve();
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 maxn = 3e5 + 5;
const long long maxc = 2e9 + 7;
const int LOG = 30;
int n, a[maxn];
struct Node {
  Node *child[2];
  vector<int> *id;
  Node() {
    memset(child, (0), sizeof(child));
    id = new vector<int>();
  }
};
Node *root = new Node();
void add(int x, int pos) {
  Node *p = root;
  for (int i = (LOG); i >= (0); --i) {
    int k = (((x) >> (i)) & 1);
    if (p->child[k] == 0) p->child[k] = new Node();
    p = p->child[k];
    p->id->push_back(pos);
  }
}
long long dp[LOG + 1][2];
void go(Node *root, int level = LOG) {
  if (level < 0) return;
  if (root->child[0] != NULL) go(root->child[0], level - 1);
  if (root->child[1] != NULL) go(root->child[1], level - 1);
  if (root->child[0] == NULL || root->child[1] == NULL) return;
  long long res = 0;
  for (auto x : *root->child[0]->id) {
    int t = lower_bound((*root->child[1]->id).begin(),
                        (*root->child[1]->id).end(), x) -
            root->child[1]->id->begin() - 1;
    res += 1LL + t;
  }
  dp[level][0] += res;
  dp[level][1] +=
      1LL * (root->child[0]->id->size()) * (root->child[1]->id->size()) - res;
}
void solve() {
  go(root);
  long long ans = 0;
  int res = 0;
  for (int level = (0); level <= (LOG); ++level) {
    ans += 1LL * min(dp[level][0], dp[level][1]);
    if (dp[level][1] < dp[level][0]) res = res ^ (1ll << (level));
  }
  cout << ans << " " << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) add(a[i], i);
  solve();
  return 0;
}

```