#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> v, idx;
long long ans = 0, inv = 0;
long long cnt[30][2];
void f(vector<int>& P, int b) {
  vector<int> L, R;
  int c1 = 0, c2 = 0;
  if (b == -1 || P.empty()) return;
  for (auto i : P) {
    if (v[i] & (1 << b)) {
      c1++;
      cnt[b][1] += c2;
      L.push_back(i);
    } else {
      c2++;
      cnt[b][0] += c1;
      R.push_back(i);
    }
  }
  f(L, b - 1);
  f(R, b - 1);
}
void solve() {
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i], idx.push_back(i);
  f(idx, 29);
  for (int i = 29; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) ans |= 1 << i;
  }
  cout << inv << ' ' << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  if (0) {
    int T;
    cin >> T;
    while (T--) solve();
  } else {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
int n;
vector<int> v, idx;
long long ans = 0, inv = 0;
long long cnt[30][2];
void f(vector<int>& P, int b) {
  vector<int> L, R;
  int c1 = 0, c2 = 0;
  if (b == -1 || P.empty()) return;
  for (auto i : P) {
    if (v[i] & (1 << b)) {
      c1++;
      cnt[b][1] += c2;
      L.push_back(i);
    } else {
      c2++;
      cnt[b][0] += c1;
      R.push_back(i);
    }
  }
  f(L, b - 1);
  f(R, b - 1);
}
void solve() {
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i], idx.push_back(i);
  f(idx, 29);
  for (int i = 29; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) ans |= 1 << i;
  }
  cout << inv << ' ' << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  if (0) {
    int T;
    cin >> T;
    while (T--) solve();
  } else {
    solve();
  }
  return 0;
}

```