#include <bits/stdc++.h>
using namespace std;
long long op[30][2];
void go(vector<int>& a, int b) {
  if (b == -1 || a.empty()) return;
  int ls = 0, os = 0;
  long long inv = 0, inv1 = 0;
  vector<int> g[2];
  for (int& i : a) {
    g[i >> b & 1].push_back(i);
    os += !(i >> b & 1);
    ls += (i >> b & 1);
    if (i >> b & 1)
      inv1 += os;
    else
      inv += ls;
  }
  op[b][0] += inv;
  op[b][1] += inv1;
  go(g[0], b - 1);
  go(g[1], b - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int& i : a) scanf("%d", &i);
  go(a, 29);
  int x = 0;
  long long inv = 0;
  for (int b = 29; ~b; b--) {
    inv += min(op[b][0], op[b][1]);
    if (op[b][0] > op[b][1]) x |= 1 << b;
  }
  printf("%I64d %d\n", inv, x);
}

### 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;
long long op[30][2];
void go(vector<int>& a, int b) {
  if (b == -1 || a.empty()) return;
  int ls = 0, os = 0;
  long long inv = 0, inv1 = 0;
  vector<int> g[2];
  for (int& i : a) {
    g[i >> b & 1].push_back(i);
    os += !(i >> b & 1);
    ls += (i >> b & 1);
    if (i >> b & 1)
      inv1 += os;
    else
      inv += ls;
  }
  op[b][0] += inv;
  op[b][1] += inv1;
  go(g[0], b - 1);
  go(g[1], b - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int& i : a) scanf("%d", &i);
  go(a, 29);
  int x = 0;
  long long inv = 0;
  for (int b = 29; ~b; b--) {
    inv += min(op[b][0], op[b][1]);
    if (op[b][0] > op[b][1]) x |= 1 << b;
  }
  printf("%I64d %d\n", inv, x);
}

```