#include <bits/stdc++.h>
using namespace std;
vector<int> q;
int n, x;
long long dp[32][2];
void rec(int bit, vector<int> v) {
  if (bit < 0 || v.empty()) return;
  vector<int> v0, v1;
  int zero = 0, one = 0;
  for (int i : v) {
    if (i & (1 << bit)) {
      v1.push_back(i);
      ++one;
      dp[bit][1] += zero;
    } else {
      v0.push_back(i);
      ++zero;
      dp[bit][0] += one;
    }
  }
  rec(bit - 1, v0);
  rec(bit - 1, v1);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int qq;
    scanf("%d", &qq);
    q.push_back(qq);
  }
  rec(29, q);
  long long tot = 0;
  for (int i = 0; i <= 29; ++i) {
    if (dp[i][0] <= dp[i][1])
      tot += dp[i][0];
    else
      x |= (1 << i), tot += dp[i][1];
  }
  printf("%lld %d\n", tot, x);
  return 0;
}

### Prompt

Develop a solution in cpp 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;
vector<int> q;
int n, x;
long long dp[32][2];
void rec(int bit, vector<int> v) {
  if (bit < 0 || v.empty()) return;
  vector<int> v0, v1;
  int zero = 0, one = 0;
  for (int i : v) {
    if (i & (1 << bit)) {
      v1.push_back(i);
      ++one;
      dp[bit][1] += zero;
    } else {
      v0.push_back(i);
      ++zero;
      dp[bit][0] += one;
    }
  }
  rec(bit - 1, v0);
  rec(bit - 1, v1);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int qq;
    scanf("%d", &qq);
    q.push_back(qq);
  }
  rec(29, q);
  long long tot = 0;
  for (int i = 0; i <= 29; ++i) {
    if (dp[i][0] <= dp[i][1])
      tot += dp[i][0];
    else
      x |= (1 << i), tot += dp[i][1];
  }
  printf("%lld %d\n", tot, x);
  return 0;
}

```