#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
long long inv[30][2];
void rec(vector<long long> &a, int bit) {
  int n = int(a.size());
  if (bit == -1 || n == 0) return;
  vector<long long> ones, zeros;
  int obz = 0, zbo = 0, o = 0, z = 0;
  for (int i = 0; i < n; i++) {
    if ((a[i] >> bit) & 1) {
      o++;
      inv[bit][1] += z;
      ones.push_back(a[i]);
    } else {
      z++;
      inv[bit][0] += o;
      zeros.push_back(a[i]);
    }
  }
  rec(ones, bit - 1);
  rec(zeros, bit - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long x = 0;
  long long invs = 0;
  rec(a, 30);
  for (int bit = 29; bit >= 0; bit--) {
    if (inv[bit][1] < inv[bit][0]) {
      invs += inv[bit][1];
      x |= (1ll << bit);
    } else {
      invs += inv[bit][0];
    }
  }
  cout << invs << " " << x << endl;
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tt = 1;
  for (int tc = 1; tc <= tt; tc++) {
    solve();
  }
}

### Prompt

Develop a solution in CPP 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 INF = 1e9 + 7;
long long inv[30][2];
void rec(vector<long long> &a, int bit) {
  int n = int(a.size());
  if (bit == -1 || n == 0) return;
  vector<long long> ones, zeros;
  int obz = 0, zbo = 0, o = 0, z = 0;
  for (int i = 0; i < n; i++) {
    if ((a[i] >> bit) & 1) {
      o++;
      inv[bit][1] += z;
      ones.push_back(a[i]);
    } else {
      z++;
      inv[bit][0] += o;
      zeros.push_back(a[i]);
    }
  }
  rec(ones, bit - 1);
  rec(zeros, bit - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long x = 0;
  long long invs = 0;
  rec(a, 30);
  for (int bit = 29; bit >= 0; bit--) {
    if (inv[bit][1] < inv[bit][0]) {
      invs += inv[bit][1];
      x |= (1ll << bit);
    } else {
      invs += inv[bit][0];
    }
  }
  cout << invs << " " << x << endl;
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int tt = 1;
  for (int tc = 1; tc <= tt; tc++) {
    solve();
  }
}

```