#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
int n;
long long ans = 0, res = 0;
long long a[N], s1[N], s2[N], b[N];
long long x[N], y[N];
void solve(int l, int r, int k) {
  if (k < 0 || l > r) return;
  s1[l - 1] = s2[l - 1] = 0;
  int cnt = 0, tot = l;
  for (int i = l; i <= r; i++) {
    s1[i] = s1[i - 1];
    s2[i] = s2[i - 1];
    if (a[i] >> k & 1)
      s1[i]++, y[k] += s2[i - 1];
    else
      s2[i]++, x[k] += s1[i - 1];
  }
  for (int i = l; i <= r; i++)
    if (a[i] >> k & 1) b[tot++] = a[i];
  int mid = tot - 1;
  for (int i = l; i <= r; i++)
    if (!(a[i] >> k & 1)) b[tot++] = a[i];
  for (int i = l; i <= r; i++) a[i] = b[i];
  solve(l, mid, k - 1);
  solve(mid + 1, r, k - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve(1, n, 30);
  for (int k = 0; k <= 30; k++) {
    if (x[k] > y[k]) {
      ans += 1 << k;
    }
    res += min(x[k], y[k]);
  }
  printf("%lld %lld\n", res, ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 + 10;
int n;
long long ans = 0, res = 0;
long long a[N], s1[N], s2[N], b[N];
long long x[N], y[N];
void solve(int l, int r, int k) {
  if (k < 0 || l > r) return;
  s1[l - 1] = s2[l - 1] = 0;
  int cnt = 0, tot = l;
  for (int i = l; i <= r; i++) {
    s1[i] = s1[i - 1];
    s2[i] = s2[i - 1];
    if (a[i] >> k & 1)
      s1[i]++, y[k] += s2[i - 1];
    else
      s2[i]++, x[k] += s1[i - 1];
  }
  for (int i = l; i <= r; i++)
    if (a[i] >> k & 1) b[tot++] = a[i];
  int mid = tot - 1;
  for (int i = l; i <= r; i++)
    if (!(a[i] >> k & 1)) b[tot++] = a[i];
  for (int i = l; i <= r; i++) a[i] = b[i];
  solve(l, mid, k - 1);
  solve(mid + 1, r, k - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve(1, n, 30);
  for (int k = 0; k <= 30; k++) {
    if (x[k] > y[k]) {
      ans += 1 << k;
    }
    res += min(x[k], y[k]);
  }
  printf("%lld %lld\n", res, ans);
  return 0;
}

```