#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> a;
long long int cnt[33][2];
long long int inv, amid;
bool added[33];
void solve(vector<int> v, int bit) {
  if (bit == -1 || v.empty()) return;
  vector<int> v0, v1;
  int b0 = 0, b1 = 0;
  for (auto x : v) {
    bool f = bool(x & (1 << bit));
    if (f) {
      b1++;
      cnt[bit][0] += b0;
    } else {
      b0++;
      cnt[bit][1] += b1;
    }
    if (f)
      v1.push_back(x);
    else
      v0.push_back(x);
  }
  solve(v0, bit - 1);
  solve(v1, bit - 1);
}
void test_case() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  solve(a, 30);
  for (int i = 0; i < 30; i++) {
    if (cnt[i][0] < cnt[i][1]) amid += (1 << i);
    inv += min(cnt[i][0], cnt[i][1]);
  }
  cout << inv << " " << amid << '\n';
}
int main() {
  int t = 1;
  while (t--) test_case();
}

### Prompt

Your challenge is to write 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;
int n;
vector<int> a;
long long int cnt[33][2];
long long int inv, amid;
bool added[33];
void solve(vector<int> v, int bit) {
  if (bit == -1 || v.empty()) return;
  vector<int> v0, v1;
  int b0 = 0, b1 = 0;
  for (auto x : v) {
    bool f = bool(x & (1 << bit));
    if (f) {
      b1++;
      cnt[bit][0] += b0;
    } else {
      b0++;
      cnt[bit][1] += b1;
    }
    if (f)
      v1.push_back(x);
    else
      v0.push_back(x);
  }
  solve(v0, bit - 1);
  solve(v1, bit - 1);
}
void test_case() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  solve(a, 30);
  for (int i = 0; i < 30; i++) {
    if (cnt[i][0] < cnt[i][1]) amid += (1 << i);
    inv += min(cnt[i][0], cnt[i][1]);
  }
  cout << inv << " " << amid << '\n';
}
int main() {
  int t = 1;
  while (t--) test_case();
}

```