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;
struct Node {
  int down;
  int left, right;
  Node() : down(0), left(-1), right(-1) {}
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  while (cin >> n) {
    int a;
    vector<Node> tree(1);
    vector<long long> left(32), right(32);
    for (int i = 0; i < n; i++) {
      cin >> a;
      int cur = 0;
      for (int pos = 31; pos >= 0; pos--) {
        if (a & (1 << pos)) {
          if (tree[cur].right == -1) {
            tree[cur].right = tree.size();
            tree.emplace_back();
          }
          right[pos] += (tree[cur].left != -1 ? tree[tree[cur].left].down : 0);
          cur = tree[cur].right;
        } else {
          if (tree[cur].left == -1) {
            tree[cur].left = tree.size();
            tree.emplace_back();
          }
          left[pos] += (tree[cur].right != -1 ? tree[tree[cur].right].down : 0);
          cur = tree[cur].left;
        }
        tree[cur].down++;
      }
    }
    int ans = 0;
    long long inv = 0;
    for (int pos = 31; pos >= 0; pos--) {
      if (left[pos] > right[pos]) {
        ans |= (1 << pos);
      }
      inv += min(left[pos], right[pos]);
    }
    cout << inv << ' ' << ans << '\n';
  }
  return 0;
}
