#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 100;
int n, a[maxn], b[maxn];
long long sum[maxn][2];
void solve(vector<int> v, int bit) {
  if (v.empty() || bit < 0) return;
  vector<int> zero, one;
  long long zeronum = 0, onenum = 0;
  for (vector<int>::iterator it = v.begin(); it != v.end(); ++it)
    if (*it & (1 << bit))
      one.push_back(*it), onenum += zero.size();
    else
      zero.push_back(*it), zeronum += one.size();
  sum[bit][0] += zeronum, sum[bit][1] += onenum;
  solve(zero, bit - 1), solve(one, bit - 1);
}
int main() {
  scanf("%d", &n);
  vector<int> v;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), v.push_back(a[i]);
  solve(v, 30);
  long long res = 0;
  int x = 0;
  for (int i = 30; i >= 0; --i)
    if (sum[i][0] <= sum[i][1])
      res += sum[i][0];
    else
      res += sum[i][1], x += (1 << i);
  printf("%lld %d\n", res, x);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int maxn = 3e5 + 100;
int n, a[maxn], b[maxn];
long long sum[maxn][2];
void solve(vector<int> v, int bit) {
  if (v.empty() || bit < 0) return;
  vector<int> zero, one;
  long long zeronum = 0, onenum = 0;
  for (vector<int>::iterator it = v.begin(); it != v.end(); ++it)
    if (*it & (1 << bit))
      one.push_back(*it), onenum += zero.size();
    else
      zero.push_back(*it), zeronum += one.size();
  sum[bit][0] += zeronum, sum[bit][1] += onenum;
  solve(zero, bit - 1), solve(one, bit - 1);
}
int main() {
  scanf("%d", &n);
  vector<int> v;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), v.push_back(a[i]);
  solve(v, 30);
  long long res = 0;
  int x = 0;
  for (int i = 30; i >= 0; --i)
    if (sum[i][0] <= sum[i][1])
      res += sum[i][0];
    else
      res += sum[i][1], x += (1 << i);
  printf("%lld %d\n", res, x);
  return 0;
}

```