#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int n, a[N];
long long mergeSort(int a, int b, int *arr) {
  if (a + 1 == b) return 0;
  int m = (a + b) / 2;
  long long inv = mergeSort(a, m, arr) + mergeSort(m, b, arr);
  int c[m - a];
  for (int i = a; i < m; i++) c[i - a] = arr[i];
  int i = a, j = m, p = a;
  while (i < m || j < b) {
    if (i == m)
      arr[p++] = arr[j++];
    else if (j == b)
      arr[p++] = c[i++ - a];
    else if (c[i - a] <= arr[j])
      arr[p++] = c[i++ - a];
    else
      inv += m - i, arr[p++] = arr[j++];
  }
  return inv;
}
long long calc(int x) {
  int b[n];
  for (int i = 0; i < n; i++) b[i] = a[i], a[i] ^= x;
  long long inv = mergeSort(0, n, a);
  for (int i = 0; i < n; i++) a[i] = b[i];
  return inv;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int x = 0;
  long long current = calc(0);
  for (int i = 30; i >= 0; i--) {
    int xx = x ^ (1 << i);
    long long q = calc(xx);
    if (q < current) current = q, x = xx;
  }
  cout << current << " " << x << '\n';
  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;
const int N = 3e5 + 5;
int n, a[N];
long long mergeSort(int a, int b, int *arr) {
  if (a + 1 == b) return 0;
  int m = (a + b) / 2;
  long long inv = mergeSort(a, m, arr) + mergeSort(m, b, arr);
  int c[m - a];
  for (int i = a; i < m; i++) c[i - a] = arr[i];
  int i = a, j = m, p = a;
  while (i < m || j < b) {
    if (i == m)
      arr[p++] = arr[j++];
    else if (j == b)
      arr[p++] = c[i++ - a];
    else if (c[i - a] <= arr[j])
      arr[p++] = c[i++ - a];
    else
      inv += m - i, arr[p++] = arr[j++];
  }
  return inv;
}
long long calc(int x) {
  int b[n];
  for (int i = 0; i < n; i++) b[i] = a[i], a[i] ^= x;
  long long inv = mergeSort(0, n, a);
  for (int i = 0; i < n; i++) a[i] = b[i];
  return inv;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int x = 0;
  long long current = calc(0);
  for (int i = 30; i >= 0; i--) {
    int xx = x ^ (1 << i);
    long long q = calc(xx);
    if (q < current) current = q, x = xx;
  }
  cout << current << " " << x << '\n';
  return 0;
}

```