#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long> > groups;
  groups.push_back(a);
  long long x = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; i--) {
    long long zero = 0, one = 0;
    vector<vector<long long> > ngroups;
    for (const vector<long long> &group : groups) {
      long long t_inv = 0;
      long long n_inv = 0;
      long long sz = group.size();
      long long invs = (sz * (sz - 1)) / 2;
      long long bi = 0;
      long long nbi = 0;
      vector<long long> g0, g1;
      for (auto el : group) {
        if (el & (1ll << i)) {
          bi++;
          n_inv += nbi;
          g1.push_back(el);
        } else {
          nbi++;
          t_inv += bi;
          g0.push_back(el);
        }
      }
      if (!g0.empty()) ngroups.push_back(g0);
      if (!g1.empty()) ngroups.push_back(g1);
      zero += t_inv;
      one += n_inv;
    }
    if (zero > one) x |= (1ll << i);
    inv += min(one, zero);
    groups = ngroups;
  }
  cout << inv << ' ' << x << '\n';
}

### 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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long> > groups;
  groups.push_back(a);
  long long x = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; i--) {
    long long zero = 0, one = 0;
    vector<vector<long long> > ngroups;
    for (const vector<long long> &group : groups) {
      long long t_inv = 0;
      long long n_inv = 0;
      long long sz = group.size();
      long long invs = (sz * (sz - 1)) / 2;
      long long bi = 0;
      long long nbi = 0;
      vector<long long> g0, g1;
      for (auto el : group) {
        if (el & (1ll << i)) {
          bi++;
          n_inv += nbi;
          g1.push_back(el);
        } else {
          nbi++;
          t_inv += bi;
          g0.push_back(el);
        }
      }
      if (!g0.empty()) ngroups.push_back(g0);
      if (!g1.empty()) ngroups.push_back(g1);
      zero += t_inv;
      one += n_inv;
    }
    if (zero > one) x |= (1ll << i);
    inv += min(one, zero);
    groups = ngroups;
  }
  cout << inv << ' ' << x << '\n';
}

```