#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
int n;
long long int arr[MAXN];
long long int inv[35], invC[35];
void f(int i, int j, int bit) {
  if (i > j || bit < 0) return;
  int ceros = 0, unos = 0;
  long long int _inv = 0, _invC = 0;
  vector<long long int> zeroA, oneA;
  for (int l = i; l <= j; ++l) {
    int al = (arr[l] & (1ll << bit)) ? 1 : 0;
    if (al == 0) {
      _inv += unos;
      zeroA.push_back(arr[l]);
      ceros++;
    } else {
      _invC += ceros;
      oneA.push_back(arr[l]);
      unos++;
    }
  }
  inv[bit] += _inv;
  invC[bit] += _invC;
  for (int l = 0; l < zeroA.size(); ++l) {
    arr[i + l] = zeroA[l];
  }
  for (int r = 0; r < oneA.size(); ++r) {
    arr[i + zeroA.size() + r] = oneA[r];
  }
  f(i, i + zeroA.size() - 1, bit - 1);
  f(i + zeroA.size(), j, bit - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  f(0, n - 1, 34);
  long long int ans = 0;
  long long int invTotales = 0;
  for (int bit = 30; bit >= 0; --bit) {
    if (invC[bit] < inv[bit]) ans |= 1ll << bit;
    invTotales += min(inv[bit], invC[bit]);
  }
  cout << invTotales << " " << ans << "\n";
  return 0;
}

### 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;
const int MAXN = 300000;
int n;
long long int arr[MAXN];
long long int inv[35], invC[35];
void f(int i, int j, int bit) {
  if (i > j || bit < 0) return;
  int ceros = 0, unos = 0;
  long long int _inv = 0, _invC = 0;
  vector<long long int> zeroA, oneA;
  for (int l = i; l <= j; ++l) {
    int al = (arr[l] & (1ll << bit)) ? 1 : 0;
    if (al == 0) {
      _inv += unos;
      zeroA.push_back(arr[l]);
      ceros++;
    } else {
      _invC += ceros;
      oneA.push_back(arr[l]);
      unos++;
    }
  }
  inv[bit] += _inv;
  invC[bit] += _invC;
  for (int l = 0; l < zeroA.size(); ++l) {
    arr[i + l] = zeroA[l];
  }
  for (int r = 0; r < oneA.size(); ++r) {
    arr[i + zeroA.size() + r] = oneA[r];
  }
  f(i, i + zeroA.size() - 1, bit - 1);
  f(i + zeroA.size(), j, bit - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  f(0, n - 1, 34);
  long long int ans = 0;
  long long int invTotales = 0;
  for (int bit = 30; bit >= 0; --bit) {
    if (invC[bit] < inv[bit]) ans |= 1ll << bit;
    invTotales += min(inv[bit], invC[bit]);
  }
  cout << invTotales << " " << ans << "\n";
  return 0;
}

```