#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll Impact(const vector<int>& B) {
  int N = B.size();
  vector<int> R(N);
  R[N - 1] = (B[N - 1] == 0) ? 1 : 0;
  for (int i = N - 2; i >= 0; --i) {
    R[i] = R[i + 1];
    if (B[i] == 0) {
      ++R[i];
    }
  }
  ll result = 0;
  for (int i = 0; i < (int)(N - 1); ++i) {
    if (B[i] == 1) {
      result += R[i + 1];
    }
  }
  return result;
}
void Visit(vector<vector<int>>&& A, int mask, int& X, ll& result) {
  if (mask == 0) {
    return;
  }
  ll impact1 = 0, impact2 = 0;
  vector<vector<int>> A2;
  for (vector<int>& a : A) {
    vector<int> B(a.size());
    vector<int> left, right;
    for (int i = 0; i < (int)(a.size()); ++i) {
      if (a[i] & mask) {
        B[i] = 1;
        left.push_back(a[i]);
      } else {
        B[i] = 0;
        right.push_back(a[i]);
      }
    }
    impact1 += Impact(B);
    for (int i = 0; i < (int)(B.size()); ++i) {
      B[i] = 1 - B[i];
    }
    impact2 += Impact(B);
    if (!left.empty()) {
      A2.push_back(move(left));
    }
    if (!right.empty()) {
      A2.push_back(move(right));
    }
  }
  if (impact1 <= impact2) {
    result += impact1;
  } else {
    result += impact2;
    X |= mask;
  }
  Visit(move(A2), mask / 2, X, result);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<vector<int>> A(1, vector<int>(N));
  for (int& a : A[0]) {
    cin >> a;
  }
  int X = 0;
  ll result = 0;
  Visit(move(A), 1 << 30, X, result);
  cout << result << ' ' << X << '\n';
  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;
using ll = long long;
ll Impact(const vector<int>& B) {
  int N = B.size();
  vector<int> R(N);
  R[N - 1] = (B[N - 1] == 0) ? 1 : 0;
  for (int i = N - 2; i >= 0; --i) {
    R[i] = R[i + 1];
    if (B[i] == 0) {
      ++R[i];
    }
  }
  ll result = 0;
  for (int i = 0; i < (int)(N - 1); ++i) {
    if (B[i] == 1) {
      result += R[i + 1];
    }
  }
  return result;
}
void Visit(vector<vector<int>>&& A, int mask, int& X, ll& result) {
  if (mask == 0) {
    return;
  }
  ll impact1 = 0, impact2 = 0;
  vector<vector<int>> A2;
  for (vector<int>& a : A) {
    vector<int> B(a.size());
    vector<int> left, right;
    for (int i = 0; i < (int)(a.size()); ++i) {
      if (a[i] & mask) {
        B[i] = 1;
        left.push_back(a[i]);
      } else {
        B[i] = 0;
        right.push_back(a[i]);
      }
    }
    impact1 += Impact(B);
    for (int i = 0; i < (int)(B.size()); ++i) {
      B[i] = 1 - B[i];
    }
    impact2 += Impact(B);
    if (!left.empty()) {
      A2.push_back(move(left));
    }
    if (!right.empty()) {
      A2.push_back(move(right));
    }
  }
  if (impact1 <= impact2) {
    result += impact1;
  } else {
    result += impact2;
    X |= mask;
  }
  Visit(move(A2), mask / 2, X, result);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<vector<int>> A(1, vector<int>(N));
  for (int& a : A[0]) {
    cin >> a;
  }
  int X = 0;
  ll result = 0;
  Visit(move(A), 1 << 30, X, result);
  cout << result << ' ' << X << '\n';
  return 0;
}

```