#include <bits/stdc++.h>
using namespace std;
long long a[300005];
long long tmp = 0;
long long ans = 0;
long long x = 0, ee, tt, ans1, ans2, cnt1, cnt2;
unordered_map<long long, vector<long long> > cur;
long long n;
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 30 - 1; i >= 0; i--) {
    ee = pow(2, i);
    for (long long j = 0; j < n; j++) {
      cur[tmp & a[j]].push_back(a[j] & ee);
    }
    ans1 = 0, ans2 = 0;
    for (auto j : cur) {
      cnt1 = 0, cnt2 = 0;
      for (auto k : j.second) {
        if (k)
          cnt1++, ans1 += cnt2;
        else
          cnt2++, ans2 += cnt1;
      }
    }
    if (ans1 < ans2) x += ee;
    ans += min(ans1, ans2);
    tmp += ee;
    cur.clear();
  }
  cout << ans << ' ' << x << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  cerr << "\nTime elapsed: " << setprecision(5)
       << 1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Generate 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;
long long a[300005];
long long tmp = 0;
long long ans = 0;
long long x = 0, ee, tt, ans1, ans2, cnt1, cnt2;
unordered_map<long long, vector<long long> > cur;
long long n;
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 30 - 1; i >= 0; i--) {
    ee = pow(2, i);
    for (long long j = 0; j < n; j++) {
      cur[tmp & a[j]].push_back(a[j] & ee);
    }
    ans1 = 0, ans2 = 0;
    for (auto j : cur) {
      cnt1 = 0, cnt2 = 0;
      for (auto k : j.second) {
        if (k)
          cnt1++, ans1 += cnt2;
        else
          cnt2++, ans2 += cnt1;
      }
    }
    if (ans1 < ans2) x += ee;
    ans += min(ans1, ans2);
    tmp += ee;
    cur.clear();
  }
  cout << ans << ' ' << x << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  cerr << "\nTime elapsed: " << setprecision(5)
       << 1000.0 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```