#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n;
vector<int> a;
long long inv[32][2];
void solve(vector<int> &a, int depth) {
  if (depth < 0) return;
  if (a.size() == 0) return;
  vector<int> left, right;
  long long cnt0 = 0, cnt1 = 0, sum01 = 0, sum10 = 0;
  for (int i = 0; i < a.size(); ++i) {
    if (a[i] >> depth & 1) {
      cnt1++;
      sum01 += cnt0;
      right.push_back(a[i]);
    } else {
      cnt0++;
      sum10 += cnt1;
      left.push_back(a[i]);
    }
  }
  inv[depth][0] += sum01;
  inv[depth][1] += sum10;
  solve(left, depth - 1);
  solve(right, depth - 1);
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (auto &x : a) cin >> x;
  solve(a, 31);
  long long ans = 0, x = 0;
  for (int i = 0; i <= 31; ++i) {
    if (inv[i][0] >= inv[i][1])
      ans += inv[i][1];
    else
      ans += inv[i][0], x |= (1 << i);
  }
  printf("%lld %lld\n", ans, 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;
const int maxn = 1e5 + 5;
int n;
vector<int> a;
long long inv[32][2];
void solve(vector<int> &a, int depth) {
  if (depth < 0) return;
  if (a.size() == 0) return;
  vector<int> left, right;
  long long cnt0 = 0, cnt1 = 0, sum01 = 0, sum10 = 0;
  for (int i = 0; i < a.size(); ++i) {
    if (a[i] >> depth & 1) {
      cnt1++;
      sum01 += cnt0;
      right.push_back(a[i]);
    } else {
      cnt0++;
      sum10 += cnt1;
      left.push_back(a[i]);
    }
  }
  inv[depth][0] += sum01;
  inv[depth][1] += sum10;
  solve(left, depth - 1);
  solve(right, depth - 1);
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (auto &x : a) cin >> x;
  solve(a, 31);
  long long ans = 0, x = 0;
  for (int i = 0; i <= 31; ++i) {
    if (inv[i][0] >= inv[i][1])
      ans += inv[i][1];
    else
      ans += inv[i][0], x |= (1 << i);
  }
  printf("%lld %lld\n", ans, x);
  return 0;
}

```