#include <bits/stdc++.h>
using namespace std;
vector<long long int> v[8000000];
vector<long long int> zeroarray;
vector<long long int> onearray;
long long int isthebitset(long long int x, long long int k) {
  if (x & (1 << k))
    return 1;
  else
    return 0;
}
int32_t main() {
  long long int n;
  cin >> n;
  long long int index = -1;
  long long int x;
  vector<long long int> final;
  for (long long int i = 0; i < n; i++) {
    cin >> x;
    v[index + 1].push_back(x);
  }
  long long int ans = 0;
  long long int count = 0;
  vector<long long int> final_ans;
  for (long long int i = 31; i >= 0; i--) {
    long long int oneans = 0;
    long long int zeroans = 0;
    long long int temp = index + 1;
    index = count;
    for (long long int j = temp; j <= index; j++) {
      long long int one = 0, zero = 0;
      onearray.clear();
      zeroarray.clear();
      long long int len = v[j].size();
      for (long long int k = (len - 1); k >= 0; k--) {
        if (isthebitset(v[j][k], i)) {
          one++;
          zeroans += zero;
          zeroarray.push_back(v[j][k]);
        } else {
          zero++;
          oneans += one;
          onearray.push_back(v[j][k]);
        }
      }
      if (onearray.size() > 1) {
        v[count + 1] = onearray;
        reverse(v[count + 1].begin(), v[count + 1].end());
        count++;
      }
      if (zeroarray.size() > 1) {
        v[count + 1] = zeroarray;
        reverse(v[count + 1].begin(), v[count + 1].end());
        count++;
      }
    }
    ans += min(oneans, zeroans);
    if (zeroans <= oneans) {
      final_ans.push_back(0);
    } else {
      final_ans.push_back(1);
    }
  }
  cout << ans << " ";
  long long int power = 1;
  ans = 0;
  reverse(final_ans.begin(), final_ans.end());
  for (long long int i = 0; i < final_ans.size(); i++) {
    ans += final_ans[i] * power;
    power *= 2;
  }
  cout << ans << "\n";
  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;
vector<long long int> v[8000000];
vector<long long int> zeroarray;
vector<long long int> onearray;
long long int isthebitset(long long int x, long long int k) {
  if (x & (1 << k))
    return 1;
  else
    return 0;
}
int32_t main() {
  long long int n;
  cin >> n;
  long long int index = -1;
  long long int x;
  vector<long long int> final;
  for (long long int i = 0; i < n; i++) {
    cin >> x;
    v[index + 1].push_back(x);
  }
  long long int ans = 0;
  long long int count = 0;
  vector<long long int> final_ans;
  for (long long int i = 31; i >= 0; i--) {
    long long int oneans = 0;
    long long int zeroans = 0;
    long long int temp = index + 1;
    index = count;
    for (long long int j = temp; j <= index; j++) {
      long long int one = 0, zero = 0;
      onearray.clear();
      zeroarray.clear();
      long long int len = v[j].size();
      for (long long int k = (len - 1); k >= 0; k--) {
        if (isthebitset(v[j][k], i)) {
          one++;
          zeroans += zero;
          zeroarray.push_back(v[j][k]);
        } else {
          zero++;
          oneans += one;
          onearray.push_back(v[j][k]);
        }
      }
      if (onearray.size() > 1) {
        v[count + 1] = onearray;
        reverse(v[count + 1].begin(), v[count + 1].end());
        count++;
      }
      if (zeroarray.size() > 1) {
        v[count + 1] = zeroarray;
        reverse(v[count + 1].begin(), v[count + 1].end());
        count++;
      }
    }
    ans += min(oneans, zeroans);
    if (zeroans <= oneans) {
      final_ans.push_back(0);
    } else {
      final_ans.push_back(1);
    }
  }
  cout << ans << " ";
  long long int power = 1;
  ans = 0;
  reverse(final_ans.begin(), final_ans.end());
  for (long long int i = 0; i < final_ans.size(); i++) {
    ans += final_ans[i] * power;
    power *= 2;
  }
  cout << ans << "\n";
  return 0;
}

```