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 n;
long long cnt[40][20];
long long inv, x;
void sol(vector<int> a, int pos) {
  if (a.empty()) return;
  if (pos == -1) return;
  vector<int> t, tt;
  t.clear();
  tt.clear();
  long long zero = 0, one = 0;
  long long cnt1 = 0, cnt2 = 0;
  for (auto x : a) {
    if ((x >> pos) & 1) {
      one++;
      cnt2 += zero;
      t.push_back(x);
    } else {
      zero++;
      cnt1 += one;
      tt.push_back(x);
    }
  }
  cnt[pos][0] += cnt1;
  cnt[pos][1] += cnt2;
  sol(t, pos - 1);
  sol(tt, pos - 1);
}
int main() {
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i <= n - 1; i++) cin >> a[i];
  sol(a, 30);
  for (int i = 0; i <= 30; i++) {
    if (cnt[i][0] > cnt[i][1]) x |= (1 << i);
    inv += min(cnt[i][0], cnt[i][1]);
  }
  cout << inv << ' ' << x << endl;
}
