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;
int main() {
  long long n;
  scanf("%lld", &n);
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    a.push_back(x);
  }
  long long x = 0;
  long long ans = 0;
  unordered_map<long long, long long> mapa;
  for (long long i = 29; i >= 0; i--) {
    mapa.reserve(2 * n);
    long long ones = 0;
    long long zeros = 0;
    for (long long j = 0; j < n; j++) {
      long long here = a[j];
      long long z = (1 << i) - 1;
      here |= z;
      long long xorv = (1 << i);
      if (here & (xorv)) {
        ones += mapa[here ^ xorv];
      } else {
        zeros += mapa[here ^ xorv];
      }
      mapa[here]++;
    }
    mapa.clear();
    if (ones < zeros) {
      x |= (1 << i);
    }
    ans += min(ones, zeros);
  }
  cout << ans << " " << x << endl;
}
