#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
inline long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline long long mul(long long a, long long b) {
  return (long long)((long long)a * b % mod);
}
vector<long long> v;
long long n, k;
vector<long long> ct;
long long inv(vector<long long>& a, long long s, long long e) {
  if (s >= e) return 0;
  long long mid = (s + e) / 2;
  long long ans = inv(a, s, mid) + inv(a, mid + 1, e);
  for (long long i = (s); i <= (mid); i++) {
    ans += lower_bound(a.begin() + mid + 1, a.begin() + e + 1, a[i]) -
           (a.begin() + mid + 1);
  }
  sort(a.begin() + s, a.begin() + e + 1);
  return ans;
}
void fun(vector<long long>& v, long long u) {
  if (u == 0) return;
  if (v.size() <= 1) return;
  vector<long long> bg, mg;
  for (long long i = (0); i <= (v.size() - 1); i++) {
    if (v[i] & (1 << (u - 1))) {
      mg.push_back(v[i]);
      ct[u] += (i) - (v.size() - 1 - i);
    } else
      bg.push_back(v[i]);
  }
  fun(bg, u - 1);
  fun(mg, u - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  ct.assign(31, 0);
  while (t--) {
    long long x = 0;
    cin >> n;
    vector<long long> v(n, 0);
    for (long long i = (0); i <= (n - 1); i++) {
      cin >> v[i];
    }
    fun(v, 30);
    for (long long i = (1); i <= (30); i++) {
      if (ct[i] < 0) x += 1 << (i - 1);
    }
    for (long long i = (0); i <= (n - 1); i++) {
      v[i] = v[i] ^ x;
    }
    cout << inv(v, 0, v.size() - 1) << " " << x;
  }
  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 long long mod = 1e9 + 7;
inline long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline long long mul(long long a, long long b) {
  return (long long)((long long)a * b % mod);
}
vector<long long> v;
long long n, k;
vector<long long> ct;
long long inv(vector<long long>& a, long long s, long long e) {
  if (s >= e) return 0;
  long long mid = (s + e) / 2;
  long long ans = inv(a, s, mid) + inv(a, mid + 1, e);
  for (long long i = (s); i <= (mid); i++) {
    ans += lower_bound(a.begin() + mid + 1, a.begin() + e + 1, a[i]) -
           (a.begin() + mid + 1);
  }
  sort(a.begin() + s, a.begin() + e + 1);
  return ans;
}
void fun(vector<long long>& v, long long u) {
  if (u == 0) return;
  if (v.size() <= 1) return;
  vector<long long> bg, mg;
  for (long long i = (0); i <= (v.size() - 1); i++) {
    if (v[i] & (1 << (u - 1))) {
      mg.push_back(v[i]);
      ct[u] += (i) - (v.size() - 1 - i);
    } else
      bg.push_back(v[i]);
  }
  fun(bg, u - 1);
  fun(mg, u - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  ct.assign(31, 0);
  while (t--) {
    long long x = 0;
    cin >> n;
    vector<long long> v(n, 0);
    for (long long i = (0); i <= (n - 1); i++) {
      cin >> v[i];
    }
    fun(v, 30);
    for (long long i = (1); i <= (30); i++) {
      if (ct[i] < 0) x += 1 << (i - 1);
    }
    for (long long i = (0); i <= (n - 1); i++) {
      v[i] = v[i] ^ x;
    }
    cout << inv(v, 0, v.size() - 1) << " " << x;
  }
  return 0;
}

```