#include <bits/stdc++.h>
using namespace std;
long long dp[2][30];
void go(vector<int> s, int pos) {
  if (s.size() == 0) return;
  if (pos == -1) return;
  long long ones = 0;
  long long zeroes = 0;
  long long oneinv = 0;
  long long zeroinv = 0;
  vector<int> l, r;
  for (auto &x : s) {
    if ((x >> pos) & 1) {
      ones++;
      oneinv += zeroes;
      r.push_back(x);
    } else {
      zeroes++;
      zeroinv += ones;
      l.push_back(x);
    }
  }
  dp[1][pos] += oneinv;
  dp[0][pos] += zeroinv;
  go(l, pos - 1);
  go(r, pos - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < 30; i++) {
    dp[0][i] = 0;
    dp[1][i] = 0;
  }
  for (auto &x : a) cin >> x;
  go(a, 29);
  long long ans = 0;
  int x = 0;
  for (int i = 0; i < 30; i++) {
    if (dp[1][i] < dp[0][i]) {
      ans += dp[1][i];
      x |= 1 << i;
    } else {
      ans += dp[0][i];
    }
  }
  cout << ans << " " << x << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 dp[2][30];
void go(vector<int> s, int pos) {
  if (s.size() == 0) return;
  if (pos == -1) return;
  long long ones = 0;
  long long zeroes = 0;
  long long oneinv = 0;
  long long zeroinv = 0;
  vector<int> l, r;
  for (auto &x : s) {
    if ((x >> pos) & 1) {
      ones++;
      oneinv += zeroes;
      r.push_back(x);
    } else {
      zeroes++;
      zeroinv += ones;
      l.push_back(x);
    }
  }
  dp[1][pos] += oneinv;
  dp[0][pos] += zeroinv;
  go(l, pos - 1);
  go(r, pos - 1);
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < 30; i++) {
    dp[0][i] = 0;
    dp[1][i] = 0;
  }
  for (auto &x : a) cin >> x;
  go(a, 29);
  long long ans = 0;
  int x = 0;
  for (int i = 0; i < 30; i++) {
    if (dp[1][i] < dp[0][i]) {
      ans += dp[1][i];
      x |= 1 << i;
    } else {
      ans += dp[0][i];
    }
  }
  cout << ans << " " << x << endl;
}

```