#include <bits/stdc++.h>
using namespace std;
const int maxn = 3824287 + 3 + 3e5 + 10;
vector<int> g[maxn];
struct XOR_TRIE {
  int num = 0;
  int ch[maxn][2] = {0};
  void insert(int x, int id) {
    for (int i = 31, p = 0; i >= 0; --i) {
      int k = (x >> i) & 1;
      if (!ch[p][k]) ch[p][k] = ++num;
      p = ch[p][k];
      g[p].emplace_back(id);
    }
  }
} T;
long long ans[32][2];
void dfs(int deep, int root) {
  if (deep < 0) return;
  int l = T.ch[root][0], r = T.ch[root][1];
  long long lsize = g[l].size(), rsize = g[r].size();
  long long inv = 0, p = 0;
  for (int i = 0; i < lsize; ++i) {
    while (p < rsize and g[r][p] < g[l][i]) p++;
    inv += p;
  }
  ans[deep][0] += inv;
  ans[deep][1] += lsize * rsize - inv;
  if (l) dfs(deep - 1, l);
  if (r) dfs(deep - 1, r);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    T.insert(t, i);
  }
  dfs(31, 0);
  long long x = 0, inv = 0;
  for (int i = 30; i >= 0; --i) {
    if (ans[i][0] <= ans[i][1])
      inv += ans[i][0];
    else {
      inv += ans[i][1];
      x += (1ll << i);
    }
  }
  cout << inv << ' ' << x << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int maxn = 3824287 + 3 + 3e5 + 10;
vector<int> g[maxn];
struct XOR_TRIE {
  int num = 0;
  int ch[maxn][2] = {0};
  void insert(int x, int id) {
    for (int i = 31, p = 0; i >= 0; --i) {
      int k = (x >> i) & 1;
      if (!ch[p][k]) ch[p][k] = ++num;
      p = ch[p][k];
      g[p].emplace_back(id);
    }
  }
} T;
long long ans[32][2];
void dfs(int deep, int root) {
  if (deep < 0) return;
  int l = T.ch[root][0], r = T.ch[root][1];
  long long lsize = g[l].size(), rsize = g[r].size();
  long long inv = 0, p = 0;
  for (int i = 0; i < lsize; ++i) {
    while (p < rsize and g[r][p] < g[l][i]) p++;
    inv += p;
  }
  ans[deep][0] += inv;
  ans[deep][1] += lsize * rsize - inv;
  if (l) dfs(deep - 1, l);
  if (r) dfs(deep - 1, r);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    T.insert(t, i);
  }
  dfs(31, 0);
  long long x = 0, inv = 0;
  for (int i = 30; i >= 0; --i) {
    if (ans[i][0] <= ans[i][1])
      inv += ans[i][0];
    else {
      inv += ans[i][1];
      x += (1ll << i);
    }
  }
  cout << inv << ' ' << x << endl;
  return 0;
}

```