#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
int cnt = 1, tree[maxn * 30][2], a[maxn];
long long sum[maxn * 30], f[40][2], ans1, ans2;
void insert(int x, int val, int w, string s) {
  if (w < 0) {
    sum[x]++;
    return;
  }
  int c = (val >> w) & 1;
  if (!tree[x][c]) tree[x][c] = ++cnt;
  insert(tree[x][c], val, w - 1, s);
  if (c == 0)
    f[w][0] += sum[tree[x][1]];
  else
    f[w][1] += sum[tree[x][0]];
  sum[x] = sum[tree[x][0]] + sum[tree[x][1]];
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    insert(1, a[i], 29, "");
  }
  ans1 = ans2 = 0;
  for (int i = 0; i < 30; i++) {
    if (f[i][0] > f[i][1]) ans2 |= (1 << i);
    ans1 += min(f[i][0], f[i][1]);
  }
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

### Prompt

Please formulate 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 maxn = 3e5 + 10;
int cnt = 1, tree[maxn * 30][2], a[maxn];
long long sum[maxn * 30], f[40][2], ans1, ans2;
void insert(int x, int val, int w, string s) {
  if (w < 0) {
    sum[x]++;
    return;
  }
  int c = (val >> w) & 1;
  if (!tree[x][c]) tree[x][c] = ++cnt;
  insert(tree[x][c], val, w - 1, s);
  if (c == 0)
    f[w][0] += sum[tree[x][1]];
  else
    f[w][1] += sum[tree[x][0]];
  sum[x] = sum[tree[x][0]] + sum[tree[x][1]];
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    insert(1, a[i], 29, "");
  }
  ans1 = ans2 = 0;
  for (int i = 0; i < 30; i++) {
    if (f[i][0] > f[i][1]) ans2 |= (1 << i);
    ans1 += min(f[i][0], f[i][1]);
  }
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

```