#include <bits/stdc++.h>
void solve();
int main(int argc, char* argv[]) {
  ::std::ios::sync_with_stdio(false);
  ::std::cin.tie(0);
  ::std::cout.tie(0);
  int tc = 1;
  while (tc--) solve();
  return 0;
}
using ll = long long;
using pii = std::pair<int, int>;
using vi = std::vector<int>;
using vii = std::vector<pii>;
using std::cin;
using std::cout;
using std::sort;
using std::string;
using std::vector;
const int inf = 0x3f3f3f3f;
const int MOD = 998244853;
vi t[2];
ll dp[2][30];
vector<vi> g;
int id = 1;
void add(int x, int pos) {
  int v = 0;
  for (int i = 29; i >= 0; --i) {
    int bit = (x >> i) & 1;
    if (!t[bit][v]) t[bit][v] = id++;
    v = t[bit][v];
    g[v].push_back(pos);
  }
}
void go(int v, int b = 29) {
  int l = t[0][v];
  int r = t[1][v];
  if (l) go(l, b - 1);
  if (r) go(r, b - 1);
  ll res = 0;
  int p = 0;
  for (auto x : g[l]) {
    while (p < ((int)(g[r]).size()) && g[r][p] < x) ++p;
    res += p;
  }
  dp[0][b] += res;
  dp[1][b] += ((int)(g[l]).size()) * 1LL * ((int)(g[r]).size()) - res;
}
void solve() {
  int n;
  cin >> n;
  t[0].resize(n * 32);
  t[1].resize(n * 32);
  g.resize(n * 32);
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    int x;
    cin >> x;
    add(x, i + 1);
  }
  go(0);
  ll res = 0;
  int sum = 0;
  for (int i = 0; i <= 29; ++i) {
    res += std::min(dp[0][i], dp[1][i]);
    if (dp[1][i] < dp[0][i]) sum += 1 << i;
  }
  cout << res << ' ' << sum << '\n';
}

### 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>
void solve();
int main(int argc, char* argv[]) {
  ::std::ios::sync_with_stdio(false);
  ::std::cin.tie(0);
  ::std::cout.tie(0);
  int tc = 1;
  while (tc--) solve();
  return 0;
}
using ll = long long;
using pii = std::pair<int, int>;
using vi = std::vector<int>;
using vii = std::vector<pii>;
using std::cin;
using std::cout;
using std::sort;
using std::string;
using std::vector;
const int inf = 0x3f3f3f3f;
const int MOD = 998244853;
vi t[2];
ll dp[2][30];
vector<vi> g;
int id = 1;
void add(int x, int pos) {
  int v = 0;
  for (int i = 29; i >= 0; --i) {
    int bit = (x >> i) & 1;
    if (!t[bit][v]) t[bit][v] = id++;
    v = t[bit][v];
    g[v].push_back(pos);
  }
}
void go(int v, int b = 29) {
  int l = t[0][v];
  int r = t[1][v];
  if (l) go(l, b - 1);
  if (r) go(r, b - 1);
  ll res = 0;
  int p = 0;
  for (auto x : g[l]) {
    while (p < ((int)(g[r]).size()) && g[r][p] < x) ++p;
    res += p;
  }
  dp[0][b] += res;
  dp[1][b] += ((int)(g[l]).size()) * 1LL * ((int)(g[r]).size()) - res;
}
void solve() {
  int n;
  cin >> n;
  t[0].resize(n * 32);
  t[1].resize(n * 32);
  g.resize(n * 32);
  for (int i = 0, nnnn = (n); i < nnnn; ++i) {
    int x;
    cin >> x;
    add(x, i + 1);
  }
  go(0);
  ll res = 0;
  int sum = 0;
  for (int i = 0; i <= 29; ++i) {
    res += std::min(dp[0][i], dp[1][i]);
    if (dp[1][i] < dp[0][i]) sum += 1 << i;
  }
  cout << res << ' ' << sum << '\n';
}

```