#include <bits/stdc++.h>
using namespace std;
int N;
vector<vector<pair<int, int> > > v, nv;
long long count_inversions(const vector<pair<int, int> >& v0,
                           const vector<pair<int, int> >& v1) {
  long long ans = 0;
  int pos = 0;
  for (int i = 0; i < (v0.size()); i++) {
    while (pos < (int)v1.size() && v0[i].second >= v1[pos].second) pos++;
    ans += pos;
  }
  return ans;
}
int main(int argc, char* argv[]) {
  scanf("%d", &N);
  v.push_back(vector<pair<int, int> >());
  for (int i = 0; i < (N); i++) {
    int a;
    scanf("%d", &a);
    v[0].push_back(make_pair(a, i));
  }
  long long ans = 0;
  int x = 0;
  for (int b = (30); b >= (0); b--) {
    long long inv0 = 0, inv1 = 0;
    for (int i = 0; i < (v.size()); i++) {
      vector<pair<int, int> > v0, v1;
      for (int j = 0; j < (v[i].size()); j++) {
        if (v[i][j].first & (1 << b))
          v1.push_back(v[i][j]);
        else
          v0.push_back(v[i][j]);
      }
      long long inv = count_inversions(v0, v1);
      inv0 += inv;
      inv1 += (long long)v0.size() * (long long)v1.size() - inv;
      if (v0.size()) nv.push_back(v0);
      if (v1.size()) nv.push_back(v1);
    }
    if (inv0 > inv1) x |= (1 << b);
    ans += min(inv0, inv1);
    v = nv;
    nv.clear();
  }
  cout << ans << " " << x << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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 N;
vector<vector<pair<int, int> > > v, nv;
long long count_inversions(const vector<pair<int, int> >& v0,
                           const vector<pair<int, int> >& v1) {
  long long ans = 0;
  int pos = 0;
  for (int i = 0; i < (v0.size()); i++) {
    while (pos < (int)v1.size() && v0[i].second >= v1[pos].second) pos++;
    ans += pos;
  }
  return ans;
}
int main(int argc, char* argv[]) {
  scanf("%d", &N);
  v.push_back(vector<pair<int, int> >());
  for (int i = 0; i < (N); i++) {
    int a;
    scanf("%d", &a);
    v[0].push_back(make_pair(a, i));
  }
  long long ans = 0;
  int x = 0;
  for (int b = (30); b >= (0); b--) {
    long long inv0 = 0, inv1 = 0;
    for (int i = 0; i < (v.size()); i++) {
      vector<pair<int, int> > v0, v1;
      for (int j = 0; j < (v[i].size()); j++) {
        if (v[i][j].first & (1 << b))
          v1.push_back(v[i][j]);
        else
          v0.push_back(v[i][j]);
      }
      long long inv = count_inversions(v0, v1);
      inv0 += inv;
      inv1 += (long long)v0.size() * (long long)v1.size() - inv;
      if (v0.size()) nv.push_back(v0);
      if (v1.size()) nv.push_back(v1);
    }
    if (inv0 > inv1) x |= (1 << b);
    ans += min(inv0, inv1);
    v = nv;
    nv.clear();
  }
  cout << ans << " " << x << endl;
  return 0;
}

```