#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long maxb = 30;
const long long random_hash = rnd();
struct custom_hash {
  long long operator()(long long first) const { return first ^ random_hash; }
};
unordered_map<long long, long long, custom_hash> mp[2];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    long long first;
    cin >> first;
    a[i] = first;
  }
  long long ans = 0;
  long long ans_x = 0;
  for (long long b = 0; b < maxb; b++) {
    long long cnt[2] = {0, 0};
    for (long long i = 0; i < n; i++) {
      if (a[i] & 1) {
        a[i] >>= 1;
        cnt[1] += mp[0][a[i]];
        mp[1][a[i]]++;
      } else {
        a[i] >>= 1;
        cnt[0] += mp[1][a[i]];
        mp[0][a[i]]++;
      }
    }
    mp[0].clear();
    mp[1].clear();
    ans += min(cnt[0], cnt[1]);
    if (cnt[0] > cnt[1]) {
      ans_x |= (1 << b);
    }
  }
  cout << ans << " " << ans_x << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long maxb = 30;
const long long random_hash = rnd();
struct custom_hash {
  long long operator()(long long first) const { return first ^ random_hash; }
};
unordered_map<long long, long long, custom_hash> mp[2];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    long long first;
    cin >> first;
    a[i] = first;
  }
  long long ans = 0;
  long long ans_x = 0;
  for (long long b = 0; b < maxb; b++) {
    long long cnt[2] = {0, 0};
    for (long long i = 0; i < n; i++) {
      if (a[i] & 1) {
        a[i] >>= 1;
        cnt[1] += mp[0][a[i]];
        mp[1][a[i]]++;
      } else {
        a[i] >>= 1;
        cnt[0] += mp[1][a[i]];
        mp[0][a[i]]++;
      }
    }
    mp[0].clear();
    mp[1].clear();
    ans += min(cnt[0], cnt[1]);
    if (cnt[0] > cnt[1]) {
      ans_x |= (1 << b);
    }
  }
  cout << ans << " " << ans_x << "\n";
  return 0;
}

```