#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 10;
int fenw[MAXN];
int query(int i) {
  int sum = 0;
  for (i++; i > 0; i -= i & -i) sum += fenw[i];
  return sum;
}
void increment(int i) {
  for (i++; i < MAXN; i += i & -i) fenw[i]++;
}
int query(int l, int r) { return query(r - 1) - query(l - 1); }
long long tot[30][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> sorted = a;
  sort(sorted.begin(), sorted.end());
  sorted.erase(unique(sorted.begin(), sorted.end()), sorted.end());
  for (int i = 0; i < n; i++) {
    int x = a[i];
    for (int j = 0; j < 30; j++) {
      int lo = x ^ (1 << j), hi = x ^ (1 << j);
      for (int k = 0; k < j; k++) {
        if (lo & (1 << k)) lo ^= 1 << k;
        hi |= 1 << k;
      }
      int pos_lo = distance(sorted.begin(),
                            lower_bound(sorted.begin(), sorted.end(), lo));
      int pos_hi = distance(sorted.begin(),
                            upper_bound(sorted.begin(), sorted.end(), hi));
      if (pos_lo <= pos_hi) tot[j][(x >> j) & 1] += query(pos_lo, pos_hi);
    }
    increment(
        distance(sorted.begin(), lower_bound(sorted.begin(), sorted.end(), x)));
  }
  int x = 0;
  long long sol = 0;
  for (int i = 0; i < 30; i++) {
    x ^= (tot[i][0] > tot[i][1]) << i;
    sol += min(tot[i][0], tot[i][1]);
  }
  cout << sol << ' ' << x << '\n';
  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 + 10;
int fenw[MAXN];
int query(int i) {
  int sum = 0;
  for (i++; i > 0; i -= i & -i) sum += fenw[i];
  return sum;
}
void increment(int i) {
  for (i++; i < MAXN; i += i & -i) fenw[i]++;
}
int query(int l, int r) { return query(r - 1) - query(l - 1); }
long long tot[30][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> sorted = a;
  sort(sorted.begin(), sorted.end());
  sorted.erase(unique(sorted.begin(), sorted.end()), sorted.end());
  for (int i = 0; i < n; i++) {
    int x = a[i];
    for (int j = 0; j < 30; j++) {
      int lo = x ^ (1 << j), hi = x ^ (1 << j);
      for (int k = 0; k < j; k++) {
        if (lo & (1 << k)) lo ^= 1 << k;
        hi |= 1 << k;
      }
      int pos_lo = distance(sorted.begin(),
                            lower_bound(sorted.begin(), sorted.end(), lo));
      int pos_hi = distance(sorted.begin(),
                            upper_bound(sorted.begin(), sorted.end(), hi));
      if (pos_lo <= pos_hi) tot[j][(x >> j) & 1] += query(pos_lo, pos_hi);
    }
    increment(
        distance(sorted.begin(), lower_bound(sorted.begin(), sorted.end(), x)));
  }
  int x = 0;
  long long sol = 0;
  for (int i = 0; i < 30; i++) {
    x ^= (tot[i][0] > tot[i][1]) << i;
    sol += min(tot[i][0], tot[i][1]);
  }
  cout << sol << ' ' << x << '\n';
  return 0;
}

```