#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const double err = 1e-10;
const int N = 1e5 + 5;
long long inversions = 0, x = 0, n;
long long inv[31][2];
void rec(vector<int> &V, int bit) {
  if (V.empty() || bit < 0) return;
  long long c0 = 0, c1 = 0, inv0 = 0, inv1 = 0;
  vector<int> zV, oV;
  for (int i = 0; i < V.size(); i++) {
    if (V[i] & (1 << bit)) {
      inv1 += c0;
      c1++;
      oV.push_back(V[i]);
    } else {
      inv0 += c1;
      c0++;
      zV.push_back(V[i]);
    }
  }
  inv[bit][0] += inv0;
  inv[bit][1] += inv1;
  rec(oV, bit - 1);
  rec(zV, bit - 1);
}
void solve() {
  memset(inv, 0, sizeof(inv));
  cin >> n;
  vector<int> V(n);
  for (int i = 0; i < n; i++) cin >> V[i];
  rec(V, 30);
  for (int i = 0; i < 31; i++) {
    if (inv[i][1] < inv[i][0]) x += (1 << i);
    inversions += min(inv[i][1], inv[i][0]);
  }
  cout << inversions << ' ' << x << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}

### Prompt

Please formulate 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;
const long long MOD = 1e9 + 7;
const double err = 1e-10;
const int N = 1e5 + 5;
long long inversions = 0, x = 0, n;
long long inv[31][2];
void rec(vector<int> &V, int bit) {
  if (V.empty() || bit < 0) return;
  long long c0 = 0, c1 = 0, inv0 = 0, inv1 = 0;
  vector<int> zV, oV;
  for (int i = 0; i < V.size(); i++) {
    if (V[i] & (1 << bit)) {
      inv1 += c0;
      c1++;
      oV.push_back(V[i]);
    } else {
      inv0 += c1;
      c0++;
      zV.push_back(V[i]);
    }
  }
  inv[bit][0] += inv0;
  inv[bit][1] += inv1;
  rec(oV, bit - 1);
  rec(zV, bit - 1);
}
void solve() {
  memset(inv, 0, sizeof(inv));
  cin >> n;
  vector<int> V(n);
  for (int i = 0; i < n; i++) cin >> V[i];
  rec(V, 30);
  for (int i = 0; i < 31; i++) {
    if (inv[i][1] < inv[i][0]) x += (1 << i);
    inversions += min(inv[i][1], inv[i][0]);
  }
  cout << inversions << ' ' << x << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}

```