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.
#include <bits/stdc++.h>
using namespace std;
long long int add(long long int x, long long int y) {
  return (x % 1000000007 + y % 1000000007) % 1000000007;
}
long long int mul(long long int x, long long int y) {
  return ((x % 1000000007) * (y % 1000000007)) % 1000000007;
}
long long int sub(long long int x, long long int y) {
  return (x - y + 1000000007) % 1000000007;
}
void solve() {
  long long int n, i, j, ans1 = 0, ans2 = 0;
  cin >> n;
  long long int a[n];
  long long int s[n];
  unordered_map<long long int, vector<long long int>> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    s[i] = 0;
  }
  for (long long int i = 31; i >= 0; i--) {
    m.clear();
    for (long long int j = 0; j < n; j++) {
      if (1 << i & a[j]) {
        m[s[j]].push_back(1);
        s[j] = 2 * s[j] + 1;
      } else {
        m[s[j]].push_back(0);
        s[j] = 2 * s[j] + 0;
      }
    }
    long long int e = 0, f = 0;
    for (auto v : m) {
      long long int c = 0, d = 0;
      for (auto k : v.second) {
        if (k == 1) {
          c++;
          f += d;
        } else {
          d++;
          e += c;
        }
      }
    }
    if (e > f) {
      ans2 += (1 << i);
    }
    ans1 += min(e, f);
  }
  cout << ans1 << " " << ans2;
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  return 0;
}
