#include <bits/stdc++.h>
using namespace std;
long long inversions(vector<int> &v) {
  int zeros = 0;
  long long inv = 0;
  for (int i = (int)v.size() - 1; i >= 0; i--) {
    zeros += (v[i] == 0);
    if (v[i] == 1) inv += zeros;
  }
  return inv;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> v;
  vector<int> t;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    t.push_back(x);
  }
  v.push_back(t);
  long long ans = 0;
  int x = 0;
  for (int i = 30; i >= 0; i--) {
    vector<vector<int>> vv;
    long long a = 0, b = 0;
    for (auto &go : v) {
      vector<int> tmp;
      for (auto &g : go) {
        tmp.push_back((g & (1 << i)) != 0);
      }
      a += inversions(tmp);
      for (auto &g : tmp) {
        g = (1 - g);
      }
      b += inversions(tmp);
      vector<int> t1, t2;
      for (int j = 0; j < (int)tmp.size(); j++) {
        if (tmp[j] == 0)
          t1.push_back(go[j]);
        else
          t2.push_back(go[j]);
      }
      if (!t1.empty()) vv.push_back(t1);
      if (!t2.empty()) vv.push_back(t2);
    }
    if (a <= b) {
      ans += a;
    } else {
      ans += b;
      x |= (1 << i);
    }
    v = vv;
  }
  cout << ans << " " << x;
}

### 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;
long long inversions(vector<int> &v) {
  int zeros = 0;
  long long inv = 0;
  for (int i = (int)v.size() - 1; i >= 0; i--) {
    zeros += (v[i] == 0);
    if (v[i] == 1) inv += zeros;
  }
  return inv;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int>> v;
  vector<int> t;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    t.push_back(x);
  }
  v.push_back(t);
  long long ans = 0;
  int x = 0;
  for (int i = 30; i >= 0; i--) {
    vector<vector<int>> vv;
    long long a = 0, b = 0;
    for (auto &go : v) {
      vector<int> tmp;
      for (auto &g : go) {
        tmp.push_back((g & (1 << i)) != 0);
      }
      a += inversions(tmp);
      for (auto &g : tmp) {
        g = (1 - g);
      }
      b += inversions(tmp);
      vector<int> t1, t2;
      for (int j = 0; j < (int)tmp.size(); j++) {
        if (tmp[j] == 0)
          t1.push_back(go[j]);
        else
          t2.push_back(go[j]);
      }
      if (!t1.empty()) vv.push_back(t1);
      if (!t2.empty()) vv.push_back(t2);
    }
    if (a <= b) {
      ans += a;
    } else {
      ans += b;
      x |= (1 << i);
    }
    v = vv;
  }
  cout << ans << " " << x;
}

```