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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<vector<long long> > v(1);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    v[0].push_back(x);
  }
  long long ans = 0;
  long long ind = 0;
  vector<long long> anss;
  for (long long i = 31; i >= 0; --i) {
    long long ans1 = 0;
    long long ans2 = 0;
    vector<long long> v1;
    vector<long long> v2;
    long long c = v.size();
    for (long long j = ind; j < c; ++j) {
      long long one = 0, zero = 0;
      vector<long long> t1;
      vector<long long> t2;
      for (long long k = v[j].size() - 1; k >= 0; --k) {
        if ((v[j][k] & (1 << i))) {
          ans1 += zero;
          one++;
          t1.push_back(v[j][k]);
        } else {
          ans2 += one;
          zero++;
          t2.push_back(v[j][k]);
        }
      }
      if (t1.size() > 0) {
        reverse(t1.begin(), t1.end());
        v.push_back(t1);
      }
      if (t2.size() > 0) {
        reverse(t2.begin(), t2.end());
        v.push_back(t2);
      }
    }
    ind = c;
    ans += min(ans1, ans2);
    if (ans1 <= ans2) {
      anss.push_back(0);
    } else {
      anss.push_back(1);
    }
  }
  cout << ans << " ";
  ans = 0;
  long long pw = 1;
  reverse(anss.begin(), anss.end());
  for (long long i = 0; i < anss.size(); ++i) {
    ans += anss[i] * pw;
    pw *= 2;
  }
  cout << ans;
}
