#include <bits/stdc++.h>
using namespace std;
template <typename T>
using Vec = std::vector<T>;
void solve() {
  int n;
  cin >> n;
  Vec<int> a(n);
  for (auto& ai : a) cin >> ai;
  auto cnt = Vec<Vec<int64_t>>(31, Vec<int64_t>(2));
  std::function<void(int, Vec<int>&&)> dfs = [&cnt, &dfs](int d,
                                                          Vec<int>&& vec) {
    if (d < 0 || vec.size() == 0) return;
    auto mask = 1 << d;
    Vec<int> l, r;
    auto bit0 = 0, bit1 = 0;
    for (auto i = 0u; i < vec.size(); ++i) {
      if (vec[i] & mask) {
        cnt[d][1] += bit0;
        ++bit1;
        l.push_back(vec[i]);
      } else {
        cnt[d][0] += bit1;
        ++bit0;
        r.push_back(vec[i]);
      }
    }
    dfs(d - 1, move(l));
    dfs(d - 1, move(r));
  };
  dfs(30, move(a));
  auto ans = 0LL;
  auto x = 0;
  for (int i = 0; i < 31; ++i) {
    if (cnt[i][0] <= cnt[i][1]) {
      ans += cnt[i][0];
    } else {
      ans += cnt[i][1];
      x |= (1 << i);
    }
  }
  cout << ans << " " << x << endl;
  return;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
template <typename T>
using Vec = std::vector<T>;
void solve() {
  int n;
  cin >> n;
  Vec<int> a(n);
  for (auto& ai : a) cin >> ai;
  auto cnt = Vec<Vec<int64_t>>(31, Vec<int64_t>(2));
  std::function<void(int, Vec<int>&&)> dfs = [&cnt, &dfs](int d,
                                                          Vec<int>&& vec) {
    if (d < 0 || vec.size() == 0) return;
    auto mask = 1 << d;
    Vec<int> l, r;
    auto bit0 = 0, bit1 = 0;
    for (auto i = 0u; i < vec.size(); ++i) {
      if (vec[i] & mask) {
        cnt[d][1] += bit0;
        ++bit1;
        l.push_back(vec[i]);
      } else {
        cnt[d][0] += bit1;
        ++bit0;
        r.push_back(vec[i]);
      }
    }
    dfs(d - 1, move(l));
    dfs(d - 1, move(r));
  };
  dfs(30, move(a));
  auto ans = 0LL;
  auto x = 0;
  for (int i = 0; i < 31; ++i) {
    if (cnt[i][0] <= cnt[i][1]) {
      ans += cnt[i][0];
    } else {
      ans += cnt[i][1];
      x |= (1 << i);
    }
  }
  cout << ans << " " << x << endl;
  return;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```