#include <bits/stdc++.h>
using namespace std;
inline bool valid(int x, int n) { return 0 <= x && x < n; }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
template <typename T>
inline T pop(queue<T>& q) {
  T front = q.front();
  q.pop();
  return front;
}
template <typename T>
inline T gcd(T a, T b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
template <typename T>
tuple<T, T, T> xgcd(T a, T b) {
  if (b == 0) return {1, 0, a};
  T x, y, g;
  tie(x, y, g) = xgcd(b, a % b);
  return {y, x - (a / b) * y, g};
}
long long int rst = 0;
long long int x = 0;
void solve(vector<vector<long long int> >& vv, int k) {
  if (k == -1) return;
  long long int sub0 = 0;
  long long int sub1 = 0;
  vector<vector<long long int> > next;
  for (vector<long long int>& v : vv) {
    vector<long long int> one, zero;
    int s0 = 0;
    int s1 = 0;
    int n = v.size();
    if (n == 0) continue;
    for (int i = n - 1; i >= 0; i--) {
      if (v[i] & (1ll << k)) {
        sub0 += s0;
        s1++;
        one.push_back(v[i]);
      } else {
        s0++;
        sub1 += s1;
        zero.push_back(v[i]);
      }
    }
    reverse(one.begin(), one.end());
    reverse(zero.begin(), zero.end());
    next.push_back(zero);
    next.push_back(one);
  }
  if (sub0 > sub1) {
    x |= (1 << k);
    for (int i = 0; i < next.size(); i += 2) swap(next[i], next[i + 1]);
  }
  rst += min(sub0, sub1);
  solve(next, k - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<vector<long long int> > vv{v};
  solve(vv, 30);
  cout << rst << " " << x << "\n";
}

### Prompt

Please provide a Cpp coded solution 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;
inline bool valid(int x, int n) { return 0 <= x && x < n; }
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
template <typename T>
inline T pop(queue<T>& q) {
  T front = q.front();
  q.pop();
  return front;
}
template <typename T>
inline T gcd(T a, T b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
template <typename T>
tuple<T, T, T> xgcd(T a, T b) {
  if (b == 0) return {1, 0, a};
  T x, y, g;
  tie(x, y, g) = xgcd(b, a % b);
  return {y, x - (a / b) * y, g};
}
long long int rst = 0;
long long int x = 0;
void solve(vector<vector<long long int> >& vv, int k) {
  if (k == -1) return;
  long long int sub0 = 0;
  long long int sub1 = 0;
  vector<vector<long long int> > next;
  for (vector<long long int>& v : vv) {
    vector<long long int> one, zero;
    int s0 = 0;
    int s1 = 0;
    int n = v.size();
    if (n == 0) continue;
    for (int i = n - 1; i >= 0; i--) {
      if (v[i] & (1ll << k)) {
        sub0 += s0;
        s1++;
        one.push_back(v[i]);
      } else {
        s0++;
        sub1 += s1;
        zero.push_back(v[i]);
      }
    }
    reverse(one.begin(), one.end());
    reverse(zero.begin(), zero.end());
    next.push_back(zero);
    next.push_back(one);
  }
  if (sub0 > sub1) {
    x |= (1 << k);
    for (int i = 0; i < next.size(); i += 2) swap(next[i], next[i + 1]);
  }
  rst += min(sub0, sub1);
  solve(next, k - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<vector<long long int> > vv{v};
  solve(vv, 30);
  cout << rst << " " << x << "\n";
}

```