#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const int mlog = 30;
int n;
int a[maxn];
vector<vector<int>> has[mlog + 10];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  vector<int> tmp;
  for (int i = 1; i <= n; i++) tmp.push_back(a[i]);
  has[mlog].push_back(tmp);
  long long ans = 0;
  int wow = 0;
  for (int i = mlog; i >= 0; i--) {
    long long res0 = 0, res1 = 0;
    for (auto p : has[i]) {
      int cnt0 = 0, cnt1 = 0;
      vector<int> has0, has1;
      for (auto x : p) {
        int bit = !!(x & (1 << i));
        if (bit == 0) {
          cnt0++;
          res0 += cnt1;
          has0.push_back(x);
        } else {
          cnt1++;
          res1 += cnt0;
          has1.push_back(x);
        }
      }
      if (i > 0) {
        if (!has0.empty()) has[i - 1].push_back(has0);
        if (!has1.empty()) has[i - 1].push_back(has1);
      }
    }
    ans += min(res0, res1);
    if (res0 > res1) wow ^= (1 << i);
  }
  printf("%lld %d", ans, wow);
}

### 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>
using namespace std;
const int maxn = 3e5 + 5;
const int mlog = 30;
int n;
int a[maxn];
vector<vector<int>> has[mlog + 10];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  vector<int> tmp;
  for (int i = 1; i <= n; i++) tmp.push_back(a[i]);
  has[mlog].push_back(tmp);
  long long ans = 0;
  int wow = 0;
  for (int i = mlog; i >= 0; i--) {
    long long res0 = 0, res1 = 0;
    for (auto p : has[i]) {
      int cnt0 = 0, cnt1 = 0;
      vector<int> has0, has1;
      for (auto x : p) {
        int bit = !!(x & (1 << i));
        if (bit == 0) {
          cnt0++;
          res0 += cnt1;
          has0.push_back(x);
        } else {
          cnt1++;
          res1 += cnt0;
          has1.push_back(x);
        }
      }
      if (i > 0) {
        if (!has0.empty()) has[i - 1].push_back(has0);
        if (!has1.empty()) has[i - 1].push_back(has1);
      }
    }
    ans += min(res0, res1);
    if (res0 > res1) wow ^= (1 << i);
  }
  printf("%lld %d", ans, wow);
}

```