#include <bits/stdc++.h>
using namespace std;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
const int INF = 0x3f3f3f3f;
const int maxn = 30 * 3e5 + 70;
const int mod = 998244353;
const int MAXN = 705;
int n, tot, ch[maxn][2], cnt[maxn];
long long l[40], r[40];
void add(int x) {
  int p = 0;
  for (int i = 30; i >= 0; --i) {
    int v = x >> i & 1;
    if (!ch[p][v]) ch[p][v] = ++tot;
    if (v == 1)
      l[i] += cnt[ch[p][0]];
    else
      r[i] += cnt[ch[p][1]];
    p = ch[p][v];
    ++cnt[p];
  }
}
int main() {
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    add(a[i]);
  }
  long long x = 0, ans = 0;
  for (int i = 30; i >= 0; --i) {
    if (l[i] < r[i])
      x += 1 << i, ans += l[i];
    else
      ans += r[i];
  }
  cout << ans << " " << x << endl;
}

### 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;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
const int INF = 0x3f3f3f3f;
const int maxn = 30 * 3e5 + 70;
const int mod = 998244353;
const int MAXN = 705;
int n, tot, ch[maxn][2], cnt[maxn];
long long l[40], r[40];
void add(int x) {
  int p = 0;
  for (int i = 30; i >= 0; --i) {
    int v = x >> i & 1;
    if (!ch[p][v]) ch[p][v] = ++tot;
    if (v == 1)
      l[i] += cnt[ch[p][0]];
    else
      r[i] += cnt[ch[p][1]];
    p = ch[p][v];
    ++cnt[p];
  }
}
int main() {
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    add(a[i]);
  }
  long long x = 0, ans = 0;
  for (int i = 30; i >= 0; --i) {
    if (l[i] < r[i])
      x += 1 << i, ans += l[i];
    else
      ans += r[i];
  }
  cout << ans << " " << x << endl;
}

```