#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int M = N * 30;
int tree[M][2];
vector<int> val[M];
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int idx = 0;
  auto insert = [&](int n, int id) {
    int tmp = 0;
    bitset<30> st(n);
    for (int i = 29; i >= 0; i--) {
      if (!tree[tmp][st[i]]) tree[tmp][st[i]] = ++idx;
      tmp = tree[tmp][st[i]];
      val[tmp].emplace_back(id);
    }
  };
  for (int i = 1; i <= n; i++) insert(a[i], i);
  auto get_cal = [&](vector<int> &a, vector<int> &b) -> long long {
    long long ans = 0;
    int i = 0, j = 0;
    while (i < a.size() && j < b.size()) {
      if (a[i] < b[j])
        i++;
      else
        ans += (int)a.size() - i, j++;
    }
    return ans;
  };
  vector<vector<long long>> ans(30, vector<long long>(2, 0));
  function<void(int, int)> query = [&](int root, int deep) {
    if (!tree[root][0] && !tree[root][1]) return;
    long long cnt = get_cal(val[tree[root][0]], val[tree[root][1]]);
    ans[deep][0] += cnt;
    ans[deep][1] +=
        1LL * val[tree[root][0]].size() * val[tree[root][1]].size() - cnt;
    if (tree[root][0]) query(tree[root][0], deep - 1);
    if (tree[root][1]) query(tree[root][1], deep - 1);
  };
  query(0, 29);
  int res = 0;
  long long cnt = 0;
  for (int i = 29; i >= 0; i--) {
    if (ans[i][0] > ans[i][1])
      res += 1 << i, cnt += ans[i][1];
    else
      cnt += ans[i][0];
  }
  printf("%lld %d\n", cnt, res);
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 3e5 + 10;
const int M = N * 30;
int tree[M][2];
vector<int> val[M];
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int idx = 0;
  auto insert = [&](int n, int id) {
    int tmp = 0;
    bitset<30> st(n);
    for (int i = 29; i >= 0; i--) {
      if (!tree[tmp][st[i]]) tree[tmp][st[i]] = ++idx;
      tmp = tree[tmp][st[i]];
      val[tmp].emplace_back(id);
    }
  };
  for (int i = 1; i <= n; i++) insert(a[i], i);
  auto get_cal = [&](vector<int> &a, vector<int> &b) -> long long {
    long long ans = 0;
    int i = 0, j = 0;
    while (i < a.size() && j < b.size()) {
      if (a[i] < b[j])
        i++;
      else
        ans += (int)a.size() - i, j++;
    }
    return ans;
  };
  vector<vector<long long>> ans(30, vector<long long>(2, 0));
  function<void(int, int)> query = [&](int root, int deep) {
    if (!tree[root][0] && !tree[root][1]) return;
    long long cnt = get_cal(val[tree[root][0]], val[tree[root][1]]);
    ans[deep][0] += cnt;
    ans[deep][1] +=
        1LL * val[tree[root][0]].size() * val[tree[root][1]].size() - cnt;
    if (tree[root][0]) query(tree[root][0], deep - 1);
    if (tree[root][1]) query(tree[root][1], deep - 1);
  };
  query(0, 29);
  int res = 0;
  long long cnt = 0;
  for (int i = 29; i >= 0; i--) {
    if (ans[i][0] > ans[i][1])
      res += 1 << i, cnt += ans[i][1];
    else
      cnt += ans[i][0];
  }
  printf("%lld %d\n", cnt, res);
  return 0;
}

```