#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 5;
int a[MAXN], b[MAXN], temp[MAXN];
int n, ans;
inline long long MergeSort(int l, int r) {
  if (l == r) return 0;
  int i = l, j = ((l + r) >> 1) + 1, pos = l;
  long long subans =
      MergeSort(l, ((l + r) >> 1)) + MergeSort(((l + r) >> 1) + 1, r);
  while (i <= ((l + r) >> 1) && j <= r) {
    if (a[i] <= a[j])
      temp[pos++] = a[i++];
    else
      temp[pos++] = a[j++], subans += ((l + r) >> 1) - i + 1;
  }
  while (i <= ((l + r) >> 1)) temp[pos++] = a[i++];
  while (j <= r) temp[pos++] = a[j++];
  for (int i = l; i <= r; i++) a[i] = temp[i];
  return subans;
}
inline long long InversePair(int x) {
  for (int i = 1; i <= n; i++) a[i] = b[i] ^ x;
  return MergeSort(1, n);
}
signed main(void) {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  long long base = InversePair(0);
  for (int i = 0; i <= 31; i++) {
    long long sub = InversePair(1 << i);
    if (sub < base) ans += (1 << i);
  }
  printf("%lld %d\n", InversePair(ans), ans);
  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;
const int MAXN = 3e5 + 5;
int a[MAXN], b[MAXN], temp[MAXN];
int n, ans;
inline long long MergeSort(int l, int r) {
  if (l == r) return 0;
  int i = l, j = ((l + r) >> 1) + 1, pos = l;
  long long subans =
      MergeSort(l, ((l + r) >> 1)) + MergeSort(((l + r) >> 1) + 1, r);
  while (i <= ((l + r) >> 1) && j <= r) {
    if (a[i] <= a[j])
      temp[pos++] = a[i++];
    else
      temp[pos++] = a[j++], subans += ((l + r) >> 1) - i + 1;
  }
  while (i <= ((l + r) >> 1)) temp[pos++] = a[i++];
  while (j <= r) temp[pos++] = a[j++];
  for (int i = l; i <= r; i++) a[i] = temp[i];
  return subans;
}
inline long long InversePair(int x) {
  for (int i = 1; i <= n; i++) a[i] = b[i] ^ x;
  return MergeSort(1, n);
}
signed main(void) {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", b + i);
  long long base = InversePair(0);
  for (int i = 0; i <= 31; i++) {
    long long sub = InversePair(1 << i);
    if (sub < base) ans += (1 << i);
  }
  printf("%lld %d\n", InversePair(ans), ans);
  return 0;
}

```