#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<tuple<int, int, int>> v(n);
  long long ans = 0;
  int x = 0;
  for (int bit = 0; bit < 30; bit++) {
    for (int i = 0; i < n; i++) {
      v[i] = make_tuple(a[i] >> (bit + 1), i, (a[i] >> bit) & 1);
    }
    sort(v.begin(), v.end());
    long long inv0 = 0;
    long long inv1 = 0;
    int beg = 0;
    while (beg < n) {
      int end = beg;
      while (end + 1 < n && get<0>(v[end + 1]) == get<0>(v[end])) {
        ++end;
      }
      int k0 = 0;
      int k1 = 0;
      for (int i = beg; i <= end; i++) {
        int bit = get<2>(v[i]);
        if (bit == 0) {
          inv0 += k1;
          k0 += 1;
        } else {
          inv1 += k0;
          k1 += 1;
        }
      }
      beg = end + 1;
    }
    if (inv0 <= inv1) {
      ans += inv0;
    } else {
      ans += inv1;
      x += (1 << bit);
    }
  }
  cout << ans << " " << x << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<tuple<int, int, int>> v(n);
  long long ans = 0;
  int x = 0;
  for (int bit = 0; bit < 30; bit++) {
    for (int i = 0; i < n; i++) {
      v[i] = make_tuple(a[i] >> (bit + 1), i, (a[i] >> bit) & 1);
    }
    sort(v.begin(), v.end());
    long long inv0 = 0;
    long long inv1 = 0;
    int beg = 0;
    while (beg < n) {
      int end = beg;
      while (end + 1 < n && get<0>(v[end + 1]) == get<0>(v[end])) {
        ++end;
      }
      int k0 = 0;
      int k1 = 0;
      for (int i = beg; i <= end; i++) {
        int bit = get<2>(v[i]);
        if (bit == 0) {
          inv0 += k1;
          k0 += 1;
        } else {
          inv1 += k0;
          k1 += 1;
        }
      }
      beg = end + 1;
    }
    if (inv0 <= inv1) {
      ans += inv0;
    } else {
      ans += inv1;
      x += (1 << bit);
    }
  }
  cout << ans << " " << x << '\n';
  return 0;
}

```