#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int tr[N * 30][2], idx;
vector<int> vec[N * 30];
long long f[31][2];
int a[N];
void insert(int val, int id) {
  int p = 0;
  for (int i = 30; i >= 0; i--) {
    int x = (val >> i) & 1;
    if (!tr[p][x]) tr[p][x] = ++idx;
    p = tr[p][x];
    vec[p].push_back(id);
  }
}
void dfs(int dep, int p) {
  if (dep < 0) return;
  int l = tr[p][0], r = tr[p][1];
  long long cnt = 0, idx = 0;
  for (int i = 0; i < vec[l].size(); i++) {
    while (idx < vec[r].size() && vec[l][i] > vec[r][idx]) idx++;
    cnt += idx;
  }
  f[dep][0] += cnt;
  f[dep][1] += 1LL * vec[l].size() * vec[r].size() - cnt;
  if (tr[p][0]) dfs(dep - 1, tr[p][0]);
  if (tr[p][1]) dfs(dep - 1, tr[p][1]);
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    insert(a[i], i);
  }
  dfs(30, 0);
  int num = 0;
  long long res = 0;
  for (int i = 30; i >= 0; i--) {
    if (f[i][0] <= f[i][1]) {
      res += f[i][0];
    } else {
      res += f[i][1];
      num += (1 << i);
    }
  }
  printf("%lld %d\n", res, num);
}
int main() {
  solve();
  return 0;
}

### 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 N = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int tr[N * 30][2], idx;
vector<int> vec[N * 30];
long long f[31][2];
int a[N];
void insert(int val, int id) {
  int p = 0;
  for (int i = 30; i >= 0; i--) {
    int x = (val >> i) & 1;
    if (!tr[p][x]) tr[p][x] = ++idx;
    p = tr[p][x];
    vec[p].push_back(id);
  }
}
void dfs(int dep, int p) {
  if (dep < 0) return;
  int l = tr[p][0], r = tr[p][1];
  long long cnt = 0, idx = 0;
  for (int i = 0; i < vec[l].size(); i++) {
    while (idx < vec[r].size() && vec[l][i] > vec[r][idx]) idx++;
    cnt += idx;
  }
  f[dep][0] += cnt;
  f[dep][1] += 1LL * vec[l].size() * vec[r].size() - cnt;
  if (tr[p][0]) dfs(dep - 1, tr[p][0]);
  if (tr[p][1]) dfs(dep - 1, tr[p][1]);
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    insert(a[i], i);
  }
  dfs(30, 0);
  int num = 0;
  long long res = 0;
  for (int i = 30; i >= 0; i--) {
    if (f[i][0] <= f[i][1]) {
      res += f[i][0];
    } else {
      res += f[i][1];
      num += (1 << i);
    }
  }
  printf("%lld %d\n", res, num);
}
int main() {
  solve();
  return 0;
}

```