#include <bits/stdc++.h>
long long nr_inv[30][2];
void cnt_inversions(std::vector<int> values, int bitdepth = 30) {
  if (values.empty() || bitdepth < 0) return;
  int nr0, nr1;
  nr0 = nr1 = 0;
  std::vector<int> split1, split2;
  for (auto it : values) {
    if ((1 << bitdepth) & it) {
      nr1++;
      split1.push_back(it);
      nr_inv[bitdepth][1] += nr0;
    } else {
      nr0++;
      split2.push_back(it);
      nr_inv[bitdepth][0] += nr1;
    }
  }
  cnt_inversions(split1, bitdepth - 1);
  cnt_inversions(split2, bitdepth - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  std::vector<int> values;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    values.push_back(x);
  }
  cnt_inversions(values);
  int res = 0;
  long long sum = 0LL;
  for (int i = 0; i < 30; ++i) {
    if (nr_inv[i][0] > nr_inv[i][1]) res |= (1 << i);
    sum += std::min(nr_inv[i][0], nr_inv[i][1]);
  }
  printf("%lld %d", sum, res);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
long long nr_inv[30][2];
void cnt_inversions(std::vector<int> values, int bitdepth = 30) {
  if (values.empty() || bitdepth < 0) return;
  int nr0, nr1;
  nr0 = nr1 = 0;
  std::vector<int> split1, split2;
  for (auto it : values) {
    if ((1 << bitdepth) & it) {
      nr1++;
      split1.push_back(it);
      nr_inv[bitdepth][1] += nr0;
    } else {
      nr0++;
      split2.push_back(it);
      nr_inv[bitdepth][0] += nr1;
    }
  }
  cnt_inversions(split1, bitdepth - 1);
  cnt_inversions(split2, bitdepth - 1);
}
int main() {
  int n;
  scanf("%d", &n);
  std::vector<int> values;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    values.push_back(x);
  }
  cnt_inversions(values);
  int res = 0;
  long long sum = 0LL;
  for (int i = 0; i < 30; ++i) {
    if (nr_inv[i][0] > nr_inv[i][1]) res |= (1 << i);
    sum += std::min(nr_inv[i][0], nr_inv[i][1]);
  }
  printf("%lld %d", sum, res);
  return 0;
}

```