#include <bits/stdc++.h>
using namespace std;
const int sz = 4e5;
long long a[sz], n;
long long curinv[32], flipinv[32];
void foo(vector<int> v, int b) {
  if (b < 0 || v.empty()) return;
  vector<int> zvec, ovec;
  int z = 0, o = 0;
  for (int k : v) {
    if ((k >> b) & 1) {
      o++;
      flipinv[b] += z;
      ovec.push_back(k);
    } else {
      z++;
      curinv[b] += o;
      zvec.push_back(k);
    }
  }
  foo(zvec, b - 1);
  foo(ovec, b - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  foo(v, 31);
  long long mynum = 0, totinv = 0;
  for (int b = 0; b < 32; b++) {
    if (curinv[b] <= flipinv[b])
      totinv += curinv[b];
    else {
      totinv += flipinv[b];
      mynum |= (1 << b);
    }
  }
  cout << totinv << " " << mynum << endl;
  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;
const int sz = 4e5;
long long a[sz], n;
long long curinv[32], flipinv[32];
void foo(vector<int> v, int b) {
  if (b < 0 || v.empty()) return;
  vector<int> zvec, ovec;
  int z = 0, o = 0;
  for (int k : v) {
    if ((k >> b) & 1) {
      o++;
      flipinv[b] += z;
      ovec.push_back(k);
    } else {
      z++;
      curinv[b] += o;
      zvec.push_back(k);
    }
  }
  foo(zvec, b - 1);
  foo(ovec, b - 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  foo(v, 31);
  long long mynum = 0, totinv = 0;
  for (int b = 0; b < 32; b++) {
    if (curinv[b] <= flipinv[b])
      totinv += curinv[b];
    else {
      totinv += flipinv[b];
      mynum |= (1 << b);
    }
  }
  cout << totinv << " " << mynum << endl;
  return 0;
}

```