#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 3e5 + 5;
int trie[N * 30][2];
int sz[N * 30], cnt;
long long inv[N * 30];
void insert(int x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int tmp = (x >> i) & 1;
    if (!tmp) inv[now] += sz[trie[now][!tmp]];
    if (!trie[now][tmp]) {
      trie[now][tmp] = ++cnt;
    }
    sz[trie[now][tmp]]++;
    now = trie[now][tmp];
  }
}
long long invbit[34][2];
void dfs(int now, int dep) {
  if (dep < 0) return;
  long long sum = 1ll * sz[trie[now][0]] * sz[trie[now][1]];
  invbit[dep][0] += inv[now];
  invbit[dep][1] += sum - inv[now];
  if (trie[now][0]) dfs(trie[now][0], dep - 1);
  if (trie[now][1]) dfs(trie[now][1], dep - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    insert(x);
  }
  dfs(0, 30);
  long long mincnt = 0, ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (invbit[i][0] > invbit[i][1]) {
      ans += (1 << i);
      mincnt += invbit[i][1];
    } else {
      mincnt += invbit[i][0];
    }
  }
  printf("%lld %lld\n", mincnt, ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 mod = 1e9 + 7;
const int N = 3e5 + 5;
int trie[N * 30][2];
int sz[N * 30], cnt;
long long inv[N * 30];
void insert(int x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int tmp = (x >> i) & 1;
    if (!tmp) inv[now] += sz[trie[now][!tmp]];
    if (!trie[now][tmp]) {
      trie[now][tmp] = ++cnt;
    }
    sz[trie[now][tmp]]++;
    now = trie[now][tmp];
  }
}
long long invbit[34][2];
void dfs(int now, int dep) {
  if (dep < 0) return;
  long long sum = 1ll * sz[trie[now][0]] * sz[trie[now][1]];
  invbit[dep][0] += inv[now];
  invbit[dep][1] += sum - inv[now];
  if (trie[now][0]) dfs(trie[now][0], dep - 1);
  if (trie[now][1]) dfs(trie[now][1], dep - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    insert(x);
  }
  dfs(0, 30);
  long long mincnt = 0, ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (invbit[i][0] > invbit[i][1]) {
      ans += (1 << i);
      mincnt += invbit[i][1];
    } else {
      mincnt += invbit[i][0];
    }
  }
  printf("%lld %lld\n", mincnt, ans);
  return 0;
}

```