#include <bits/stdc++.h>
using namespace std;
long long f(vector<int> a, int x, int flag) {
  int cnt = 0;
  long long s = 0;
  if (flag)
    for (int i : a) {
      if (i & (1 << x))
        s += cnt;
      else
        cnt++;
    }
  else
    for (int i : a) {
      if (i & (1 << x))
        cnt++;
      else
        s += cnt;
    }
  return s;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& i : a) cin >> i;
  long long ans = 0;
  int x = 0;
  vector<vector<int>> cur;
  cur.push_back({a});
  for (int bit = 31; bit >= 0; bit--) {
    long long v1 = 0, v2 = 0;
    vector<vector<int>> nxt;
    for (auto e : cur) {
      v1 += f(e, bit, 1);
      v2 += f(e, bit, 0);
      vector<int> A, B;
      for (int i : e) {
        if (i & (1 << bit))
          A.push_back(i);
        else
          B.push_back(i);
      }
      if ((int)(A).size()) nxt.push_back(A);
      if ((int)(B).size()) nxt.push_back(B);
    }
    cur = nxt;
    if (v1 < v2) x |= 1 << bit;
    ans += min(v1, v2);
  }
  cout << ans << ' ' << x;
  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;
long long f(vector<int> a, int x, int flag) {
  int cnt = 0;
  long long s = 0;
  if (flag)
    for (int i : a) {
      if (i & (1 << x))
        s += cnt;
      else
        cnt++;
    }
  else
    for (int i : a) {
      if (i & (1 << x))
        cnt++;
      else
        s += cnt;
    }
  return s;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& i : a) cin >> i;
  long long ans = 0;
  int x = 0;
  vector<vector<int>> cur;
  cur.push_back({a});
  for (int bit = 31; bit >= 0; bit--) {
    long long v1 = 0, v2 = 0;
    vector<vector<int>> nxt;
    for (auto e : cur) {
      v1 += f(e, bit, 1);
      v2 += f(e, bit, 0);
      vector<int> A, B;
      for (int i : e) {
        if (i & (1 << bit))
          A.push_back(i);
        else
          B.push_back(i);
      }
      if ((int)(A).size()) nxt.push_back(A);
      if ((int)(B).size()) nxt.push_back(B);
    }
    cur = nxt;
    if (v1 < v2) x |= 1 << bit;
    ans += min(v1, v2);
  }
  cout << ans << ' ' << x;
  return 0;
}

```