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.
#include <bits/stdc++.h>
using namespace std;
constexpr int INF = (sizeof(int) == 4 ? 1e9 : 2e18) + 1e5;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 2e5 + 3;
class Trie {
 private:
  static const int INT_LEN = 31;

 public:
  struct Node {
    Node *next[2] = {nullptr, nullptr};
    int id;
    vector<int> arr;
    Node(const int x) : id(x) {}
  };

 private:
  int trie_size = 0;
  int node_cnt = 0;
  Node *trie_root = this->make_node();

 private:
  Node *make_node() { return new Node(node_cnt++); }
  int trie_insert(const string &s, int pos) {
    Node *aux = this->root();
    for (const char ch : s) {
      int c = ch - '0';
      if (aux->next[c] == nullptr) aux->next[c] = this->make_node();
      aux = aux->next[c];
      aux->arr.emplace_back(pos);
    }
    ++this->trie_size;
    return aux->id;
  }

 public:
  Trie() {}
  Node *root() { return this->trie_root; }
  int insert(const int x, int pos) {
    assert(x >= 0);
    return this->trie_insert(bitset<INT_LEN>(x).to_string(), pos);
  }
};
long long dp[2][31];
void f(Trie::Node *node, int b = 30) {
  long long cnt = 0;
  if ((node->next)[0]) {
    f(node->next[0], b - 1);
    ++cnt;
  }
  if ((node->next)[1]) {
    f(node->next[1], b - 1);
    ++cnt;
  }
  if (cnt != 2) return;
  long long cur = 0;
  int j = 0;
  vector<int> &g0 = node->next[0]->arr;
  vector<int> &g1 = node->next[1]->arr;
  for (int x : g0) {
    while (j < g1.size() && g1[j] < x) ++j;
    cur += j;
  }
  dp[0][b] += cur;
  dp[1][b] += (long long)g1.size() * (int)g0.size() - cur;
}
void solve() {
  int n;
  cin >> n;
  Trie t;
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    t.insert(x, i);
  }
  f(t.root());
  long long cnt = 0, ans = 0;
  for (int i = 0, j = 29; i < 30; i++, --j) {
    cnt += min(dp[0][j], dp[1][j]);
    if (dp[1][j] < dp[0][j]) {
      ans |= (1ll << j);
    }
  }
  cout << cnt << ' ' << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int t = 1;
  while (t--) solve();
}
