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;
const int N = 3e5 + 3;
int n, arr[N];
long long sum[32][2];
struct node {
  int numOfValues;
  node* leftNode;
  node* rightNode;
  node() {
    leftNode = rightNode = NULL;
    numOfValues = 0;
  }
  void updateNumOfValues() { numOfValues++; }
  int getNumOfValues() { return numOfValues; }
  node* getRightNode() {
    if (rightNode == NULL) {
      rightNode = new node();
    }
    return rightNode;
  }
  node* getLeftNode() {
    if (leftNode == NULL) {
      leftNode = new node();
    }
    return leftNode;
  }
};
int BIT(int x, int t) { return x >> t & 1; }
struct trie {
  node* head;
  trie() { head = new node(); }
  void addElement(int x) {
    node* p = head;
    for (int i = 29; i >= 0; i--) {
      if (BIT(x, i)) {
        sum[i][1] += p->getLeftNode()->getNumOfValues();
        p = p->getRightNode();
      } else {
        sum[i][0] += p->getRightNode()->getNumOfValues();
        p = p->getLeftNode();
      }
      p->updateNumOfValues();
    }
  }
} myTree;
void solve() {
  long long resInsersion = 0;
  int resValue = 0;
  for (int i = 29; i >= 0; i--) {
    if (sum[i][0] <= sum[i][1]) {
      resInsersion += sum[i][0];
    } else {
      resInsersion += sum[i][1];
      resValue += 1 << i;
    }
  }
  cout << resInsersion << ' ' << resValue;
}
int main() {
  if (fopen("trash.inp", "r")) {
    freopen("trash.inp", "r", stdin);
    freopen("trash.out", "w", stdout);
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    myTree.addElement(x);
  }
  solve();
}
