#include <bits/stdc++.h>
const int N = 300010;
int n, a[N];
int c[N];
int mi[N];
std::vector<int> vec[N];
class Discretization {
 public:
  Discretization() { vec_.clear(); }
  void Init() { vec_.clear(); }
  void Add(int x) { vec_.push_back(x); }
  void Build() {
    std::sort(vec_.begin(), vec_.end());
    vec_.erase(std::unique(vec_.begin(), vec_.end()), vec_.end());
  }
  int Id(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (*y != x) {
      return -1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int Size() { return vec_.size(); }

 private:
  std::vector<int> vec_;
};
Discretization disc;
int main() {
  mi[0] = 1;
  for (int i = 1; i <= 31; ++i) {
    mi[i] = mi[i - 1] << 1;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    c[i] = 0;
  }
  int ans_x = 0;
  long long ans_inv = 0ll;
  for (int i = 31; i >= 0; --i) {
    disc.Init();
    for (int j = 1; j <= n; ++j) {
      c[j] = (c[j] << 1) | ((a[j] & mi[i]) ? 1 : 0);
      disc.Add(c[j]);
    }
    if (i) {
      disc.Build();
      for (int j = 1; j <= n; ++j) {
        vec[j].clear();
      }
      for (int j = 1; j <= n; ++j) {
        int id = disc.Id(c[j]);
        int num = ((a[j] & mi[i - 1]) ? 1 : 0);
        vec[id].push_back(num);
      }
      long long num_inv_sum = 0ll;
      long long num_inc_sum = 0ll;
      for (int j = 1; j <= n; ++j) {
        long long cnt1 = 0ll;
        long long num_inv = 0ll;
        long long num_inc = 0ll;
        long long cnt0 = 0ll;
        for (size_t k = 0; k < vec[j].size(); ++k) {
          if (!vec[j][k]) {
            num_inv += cnt1;
            ++cnt0;
          } else {
            num_inc += cnt0;
            ++cnt1;
          }
        }
        num_inv_sum += num_inv;
        num_inc_sum += num_inc;
      }
      if (num_inc_sum - num_inv_sum < 0) {
        ans_x |= mi[i - 1];
        ans_inv += num_inc_sum;
      } else {
        ans_inv += num_inv_sum;
      }
    }
  }
  printf("%lld %d\n", ans_inv, ans_x);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
const int N = 300010;
int n, a[N];
int c[N];
int mi[N];
std::vector<int> vec[N];
class Discretization {
 public:
  Discretization() { vec_.clear(); }
  void Init() { vec_.clear(); }
  void Add(int x) { vec_.push_back(x); }
  void Build() {
    std::sort(vec_.begin(), vec_.end());
    vec_.erase(std::unique(vec_.begin(), vec_.end()), vec_.end());
  }
  int Id(int x) {
    auto y = std::lower_bound(vec_.begin(), vec_.end(), x);
    if (*y != x) {
      return -1;
    } else {
      return y - vec_.begin() + 1;
    }
  }
  int Size() { return vec_.size(); }

 private:
  std::vector<int> vec_;
};
Discretization disc;
int main() {
  mi[0] = 1;
  for (int i = 1; i <= 31; ++i) {
    mi[i] = mi[i - 1] << 1;
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    c[i] = 0;
  }
  int ans_x = 0;
  long long ans_inv = 0ll;
  for (int i = 31; i >= 0; --i) {
    disc.Init();
    for (int j = 1; j <= n; ++j) {
      c[j] = (c[j] << 1) | ((a[j] & mi[i]) ? 1 : 0);
      disc.Add(c[j]);
    }
    if (i) {
      disc.Build();
      for (int j = 1; j <= n; ++j) {
        vec[j].clear();
      }
      for (int j = 1; j <= n; ++j) {
        int id = disc.Id(c[j]);
        int num = ((a[j] & mi[i - 1]) ? 1 : 0);
        vec[id].push_back(num);
      }
      long long num_inv_sum = 0ll;
      long long num_inc_sum = 0ll;
      for (int j = 1; j <= n; ++j) {
        long long cnt1 = 0ll;
        long long num_inv = 0ll;
        long long num_inc = 0ll;
        long long cnt0 = 0ll;
        for (size_t k = 0; k < vec[j].size(); ++k) {
          if (!vec[j][k]) {
            num_inv += cnt1;
            ++cnt0;
          } else {
            num_inc += cnt0;
            ++cnt1;
          }
        }
        num_inv_sum += num_inv;
        num_inc_sum += num_inc;
      }
      if (num_inc_sum - num_inv_sum < 0) {
        ans_x |= mi[i - 1];
        ans_inv += num_inc_sum;
      } else {
        ans_inv += num_inv_sum;
      }
    }
  }
  printf("%lld %d\n", ans_inv, ans_x);
  return 0;
}

```