#include <bits/stdc++.h>
using namespace std;
long long n;
vector<long long> a;
long long sum[2][31];
struct node {
  long long cnt;
  node* next[2];
  vector<long long> v;
  node() { memset(next, 0, sizeof next); }
};
node* root = new node();
void insert(long long x, long long idx) {
  node* cur = root;
  for (long long i = 30; i >= 0; i--) {
    long long dig = x & (1 << i) ? 1 : 0;
    if (cur->next[dig] == NULL) {
      cur->next[dig] = new node();
    }
    cur = cur->next[dig];
    cur->v.push_back(idx);
  }
}
void ser(node* n, long long idx) {
  if (idx < 0) return;
  vector<long long> v0, v1;
  if (n->next[0] != NULL) v0 = n->next[0]->v;
  if (n->next[1] != NULL) v1 = n->next[1]->v;
  if ((long long)v0.size() > 0) ser(n->next[0], idx - 1);
  if ((long long)v1.size() > 0) ser(n->next[1], idx - 1);
  if ((long long)v0.size() == 0 or (long long) v1.size() == 0) return;
  long long j = 0, tmp = 0;
  for (auto i : v0) {
    while (j < (long long)v1.size() and v1[j] < i) j++;
    tmp += j;
  }
  sum[0][idx] += tmp;
  sum[1][idx] += (long long)v0.size() * (long long)v1.size() - tmp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  while (test--) {
    cin >> n;
    a.resize(n);
    for (long long i = 0; i < (long long)n; ++i) cin >> a[i];
    for (long long i = 0; i < (long long)n; ++i) {
      insert(a[i], i);
    }
    ser(root, 30);
    long long x = 0, ans = 0;
    for (long long i = 30; i >= 0; i--) {
      long long idx = (sum[0][i] <= sum[1][i]) ? 0 : 1;
      x += (idx << i);
      ans += sum[idx][i];
    }
    cout << ans << " " << x << '\n';
  }
  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;
long long n;
vector<long long> a;
long long sum[2][31];
struct node {
  long long cnt;
  node* next[2];
  vector<long long> v;
  node() { memset(next, 0, sizeof next); }
};
node* root = new node();
void insert(long long x, long long idx) {
  node* cur = root;
  for (long long i = 30; i >= 0; i--) {
    long long dig = x & (1 << i) ? 1 : 0;
    if (cur->next[dig] == NULL) {
      cur->next[dig] = new node();
    }
    cur = cur->next[dig];
    cur->v.push_back(idx);
  }
}
void ser(node* n, long long idx) {
  if (idx < 0) return;
  vector<long long> v0, v1;
  if (n->next[0] != NULL) v0 = n->next[0]->v;
  if (n->next[1] != NULL) v1 = n->next[1]->v;
  if ((long long)v0.size() > 0) ser(n->next[0], idx - 1);
  if ((long long)v1.size() > 0) ser(n->next[1], idx - 1);
  if ((long long)v0.size() == 0 or (long long) v1.size() == 0) return;
  long long j = 0, tmp = 0;
  for (auto i : v0) {
    while (j < (long long)v1.size() and v1[j] < i) j++;
    tmp += j;
  }
  sum[0][idx] += tmp;
  sum[1][idx] += (long long)v0.size() * (long long)v1.size() - tmp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  while (test--) {
    cin >> n;
    a.resize(n);
    for (long long i = 0; i < (long long)n; ++i) cin >> a[i];
    for (long long i = 0; i < (long long)n; ++i) {
      insert(a[i], i);
    }
    ser(root, 30);
    long long x = 0, ans = 0;
    for (long long i = 30; i >= 0; i--) {
      long long idx = (sum[0][i] <= sum[1][i]) ? 0 : 1;
      x += (idx << i);
      ans += sum[idx][i];
    }
    cout << ans << " " << x << '\n';
  }
  return 0;
}

```