#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  scanf("%d", &N);
  vector<int> A(N);
  for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
  int res = 0;
  long long score = 0;
  vector<vector<int>> div(1, A);
  for (int d = 30; d >= 0; --d) {
    vector<vector<int>> nex;
    int mask = 1 << d;
    long long ok = 0, ng = 0;
    for (const auto &v : div) {
      vector<int> zero, one;
      for (auto val : v) {
        if (val & mask) {
          ok += zero.size();
          one.push_back(val);
        } else {
          ng += one.size();
          zero.push_back(val);
        }
      }
      if (!one.empty()) nex.emplace_back(one);
      if (!zero.empty()) nex.emplace_back(zero);
    }
    swap(div, nex);
    score += min(ok, ng);
    if (ok < ng) res += mask;
  }
  printf("%lld %d\n", score, res);
}

### Prompt

Your challenge is to write 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;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  scanf("%d", &N);
  vector<int> A(N);
  for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
  int res = 0;
  long long score = 0;
  vector<vector<int>> div(1, A);
  for (int d = 30; d >= 0; --d) {
    vector<vector<int>> nex;
    int mask = 1 << d;
    long long ok = 0, ng = 0;
    for (const auto &v : div) {
      vector<int> zero, one;
      for (auto val : v) {
        if (val & mask) {
          ok += zero.size();
          one.push_back(val);
        } else {
          ng += one.size();
          zero.push_back(val);
        }
      }
      if (!one.empty()) nex.emplace_back(one);
      if (!zero.empty()) nex.emplace_back(zero);
    }
    swap(div, nex);
    score += min(ok, ng);
    if (ok < ng) res += mask;
  }
  printf("%lld %d\n", score, res);
}

```