#include <bits/stdc++.h>
using namespace std;
long long n, x, y, sol;
vector<long long> v;
long long arr[32][2];
void rek(long long bit, vector<long long> a) {
  if (bit == -1 || a.size() == 0) {
    return;
  }
  vector<long long> zero;
  vector<long long> one;
  long long m = a.size();
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      one.push_back(a[i]);
    } else {
      zero.push_back(a[i]);
    }
  }
  long long cnt = 0;
  long long ostalo = 0;
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      ostalo++;
    } else {
      cnt += ostalo;
    }
  }
  arr[bit][1] += cnt;
  cnt = 0;
  ostalo = 0;
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      cnt += ostalo;
    } else {
      ostalo++;
    }
  }
  arr[bit][0] += cnt;
  rek(bit - 1, zero);
  rek(bit - 1, one);
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> y;
    v.push_back(y);
  }
  rek(31, v);
  x = 0;
  sol = 0;
  for (long long i = 0; i < 30; i++) {
    if (arr[i][1] > arr[i][0]) {
      x += (1 << i);
    }
    sol += min(arr[i][0], arr[i][1]);
  }
  cout << sol << " " << x << "\n";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, x, y, sol;
vector<long long> v;
long long arr[32][2];
void rek(long long bit, vector<long long> a) {
  if (bit == -1 || a.size() == 0) {
    return;
  }
  vector<long long> zero;
  vector<long long> one;
  long long m = a.size();
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      one.push_back(a[i]);
    } else {
      zero.push_back(a[i]);
    }
  }
  long long cnt = 0;
  long long ostalo = 0;
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      ostalo++;
    } else {
      cnt += ostalo;
    }
  }
  arr[bit][1] += cnt;
  cnt = 0;
  ostalo = 0;
  for (long long i = 0; i < m; i++) {
    if ((1 << bit) & a[i]) {
      cnt += ostalo;
    } else {
      ostalo++;
    }
  }
  arr[bit][0] += cnt;
  rek(bit - 1, zero);
  rek(bit - 1, one);
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> y;
    v.push_back(y);
  }
  rek(31, v);
  x = 0;
  sol = 0;
  for (long long i = 0; i < 30; i++) {
    if (arr[i][1] > arr[i][0]) {
      x += (1 << i);
    }
    sol += min(arr[i][0], arr[i][1]);
  }
  cout << sol << " " << x << "\n";
  return 0;
}

```