#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int num = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    num = (num << 1) + (num << 3) + (c ^ 48);
  return num;
}
vector<int> a;
long long dp[40][2];
inline void solve(vector<int> s, int p) {
  if (s.empty() || p < 0) return;
  vector<int> L, R;
  int cnt1 = 0, cnt2 = 0;
  long long ans1 = 0, ans2 = 0;
  for (int i = 0; i < s.size(); ++i) {
    int x = s[i];
    if ((x >> p) & 1) {
      ans2 += cnt1;
      ++cnt2;
      R.push_back(x);
    } else {
      ans1 += cnt2;
      ++cnt1;
      L.push_back(x);
    }
  }
  dp[p][0] += ans2;
  dp[p][1] += ans1;
  solve(L, p - 1);
  solve(R, p - 1);
}
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    a.push_back(x);
  }
  reverse(a.begin(), a.end());
  solve(a, 30);
  long long Ans = 0;
  int res = 0;
  for (int i = 30; ~i; --i) {
    if (dp[i][0] > dp[i][1])
      Ans += dp[i][1], res ^= 1 << i;
    else
      Ans += dp[i][0];
  }
  printf("%lld %d", Ans, res);
  return 0;
}

### Prompt

Please formulate 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;
inline int read() {
  int num = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    num = (num << 1) + (num << 3) + (c ^ 48);
  return num;
}
vector<int> a;
long long dp[40][2];
inline void solve(vector<int> s, int p) {
  if (s.empty() || p < 0) return;
  vector<int> L, R;
  int cnt1 = 0, cnt2 = 0;
  long long ans1 = 0, ans2 = 0;
  for (int i = 0; i < s.size(); ++i) {
    int x = s[i];
    if ((x >> p) & 1) {
      ans2 += cnt1;
      ++cnt2;
      R.push_back(x);
    } else {
      ans1 += cnt2;
      ++cnt1;
      L.push_back(x);
    }
  }
  dp[p][0] += ans2;
  dp[p][1] += ans1;
  solve(L, p - 1);
  solve(R, p - 1);
}
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    a.push_back(x);
  }
  reverse(a.begin(), a.end());
  solve(a, 30);
  long long Ans = 0;
  int res = 0;
  for (int i = 30; ~i; --i) {
    if (dp[i][0] > dp[i][1])
      Ans += dp[i][1], res ^= 1 << i;
    else
      Ans += dp[i][0];
  }
  printf("%lld %d", Ans, res);
  return 0;
}

```