#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
int sz(const T &a) {
  return int(a.size());
}
const int MAXN = 3e5 + 1;
vector<vector<int>> arr[2];
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int a;
  int cur = 0;
  arr[cur].push_back({});
  for (int i = 0; i < n; i++) {
    cin >> a;
    arr[cur].back().push_back(a);
  }
  int x = 0;
  ll inver = 0;
  vector<vector<int>> next;
  for (int i = 30; i >= 0; i--) {
    vector<ll> ans(2, 0);
    arr[!cur] = vector<vector<int>>();
    for (const auto &x : arr[cur]) {
      vector<ll> am(2, 0);
      vector<vector<int>> te;
      te.resize(2, {});
      for (auto y : x) {
        int bit = (!!(y & (1 << i)));
        if (!bit)
          ans[0] += am[1];
        else
          ans[1] += am[0];
        am[bit]++;
        te[bit].push_back(y);
      }
      for (const auto &y : te) {
        if (sz(y)) {
          arr[!cur].push_back(y);
        }
      }
    }
    if (ans[0] <= ans[1]) {
      inver += ans[0];
    } else
      inver += ans[1], x += (1 << i);
    cur = !cur;
  }
  printf("%lli %d\n", inver, x);
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
int sz(const T &a) {
  return int(a.size());
}
const int MAXN = 3e5 + 1;
vector<vector<int>> arr[2];
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int a;
  int cur = 0;
  arr[cur].push_back({});
  for (int i = 0; i < n; i++) {
    cin >> a;
    arr[cur].back().push_back(a);
  }
  int x = 0;
  ll inver = 0;
  vector<vector<int>> next;
  for (int i = 30; i >= 0; i--) {
    vector<ll> ans(2, 0);
    arr[!cur] = vector<vector<int>>();
    for (const auto &x : arr[cur]) {
      vector<ll> am(2, 0);
      vector<vector<int>> te;
      te.resize(2, {});
      for (auto y : x) {
        int bit = (!!(y & (1 << i)));
        if (!bit)
          ans[0] += am[1];
        else
          ans[1] += am[0];
        am[bit]++;
        te[bit].push_back(y);
      }
      for (const auto &y : te) {
        if (sz(y)) {
          arr[!cur].push_back(y);
        }
      }
    }
    if (ans[0] <= ans[1]) {
      inver += ans[0];
    } else
      inver += ans[1], x += (1 << i);
    cur = !cur;
  }
  printf("%lli %d\n", inver, x);
  return 0;
}

```