#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int const LOG = 31;
ll inv[LOG][3];
int what_is_here(ll val, int bit) { return (val & (1LL << bit)) != 0; }
void dq(vector<ll> &v, int bit) {
  if (bit < 0) return;
  vector<ll> down[2], seen(2, 0);
  ll one = 0, zero = 0;
  for (auto &x : v) {
    int h = what_is_here(x, bit);
    if (!h)
      zero += seen[1];
    else
      one += seen[0];
    seen[h]++;
    down[h].push_back(x);
  }
  inv[bit][0] += zero;
  inv[bit][1] += one;
  v.clear();
  if (down[0].size()) dq(down[0], bit - 1);
  if (down[1].size()) dq(down[1], bit - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<ll> v(n);
  for (auto &x : v) cin >> x;
  dq(v, 30);
  ll ans = 0, inv_ans = 0;
  for (int i = LOG - 1; i >= 0; i--) {
    if (inv[i][1] < inv[i][0]) ans ^= (1LL << i);
    inv_ans += min(inv[i][1], inv[i][0]);
  }
  cout << inv_ans << " " << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
}

### Prompt

Generate 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;
using ll = long long int;
int const LOG = 31;
ll inv[LOG][3];
int what_is_here(ll val, int bit) { return (val & (1LL << bit)) != 0; }
void dq(vector<ll> &v, int bit) {
  if (bit < 0) return;
  vector<ll> down[2], seen(2, 0);
  ll one = 0, zero = 0;
  for (auto &x : v) {
    int h = what_is_here(x, bit);
    if (!h)
      zero += seen[1];
    else
      one += seen[0];
    seen[h]++;
    down[h].push_back(x);
  }
  inv[bit][0] += zero;
  inv[bit][1] += one;
  v.clear();
  if (down[0].size()) dq(down[0], bit - 1);
  if (down[1].size()) dq(down[1], bit - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<ll> v(n);
  for (auto &x : v) cin >> x;
  dq(v, 30);
  ll ans = 0, inv_ans = 0;
  for (int i = LOG - 1; i >= 0; i--) {
    if (inv[i][1] < inv[i][0]) ans ^= (1LL << i);
    inv_ans += min(inv[i][1], inv[i][0]);
  }
  cout << inv_ans << " " << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  solve();
}

```