#include <bits/stdc++.h>
using namespace std;
long long cntInv[35][2];
void solve(vector<int>& a, int hBit) {
  if (a.size() == 0 || hBit < 0) return;
  int n = a.size();
  int cur0 = 0, cur1 = 0;
  vector<int> bit0, bit1;
  for (int i = 0; i < n; i++) {
    if (a[i] & (1 << hBit)) {
      cur1++;
      cntInv[hBit][1] += cur0;
      bit1.push_back(a[i]);
    } else {
      cur0++;
      cntInv[hBit][0] += cur1;
      bit0.push_back(a[i]);
    }
  }
  solve(bit0, hBit - 1);
  solve(bit1, hBit - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  solve(a, 30);
  int res = 0;
  long long numInv = 0;
  for (int i = 0; i <= 30; i++) {
    int ind = 0;
    if (cntInv[i][0] > cntInv[i][1]) {
      ind = 1;
      res += (1 << i);
    }
    numInv += cntInv[i][ind];
  }
  cout << numInv << " " << res << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
long long cntInv[35][2];
void solve(vector<int>& a, int hBit) {
  if (a.size() == 0 || hBit < 0) return;
  int n = a.size();
  int cur0 = 0, cur1 = 0;
  vector<int> bit0, bit1;
  for (int i = 0; i < n; i++) {
    if (a[i] & (1 << hBit)) {
      cur1++;
      cntInv[hBit][1] += cur0;
      bit1.push_back(a[i]);
    } else {
      cur0++;
      cntInv[hBit][0] += cur1;
      bit0.push_back(a[i]);
    }
  }
  solve(bit0, hBit - 1);
  solve(bit1, hBit - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  solve(a, 30);
  int res = 0;
  long long numInv = 0;
  for (int i = 0; i <= 30; i++) {
    int ind = 0;
    if (cntInv[i][0] > cntInv[i][1]) {
      ind = 1;
      res += (1 << i);
    }
    numInv += cntInv[i][ind];
  }
  cout << numInv << " " << res << endl;
  return 0;
}

```