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.
import java.io.*;
import java.util.*;

public class E {

	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	static int n;
	static int[] a;
	static int[] c1;
		
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		n = Integer.parseInt(br.readLine());
		a = new int[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		int max = 0;
		c1 = new int[n];

		for (int i=  0; i < n; i++) max = Math.max(max, c1[i] = a[i] = Integer.parseInt(st.nextToken()));
		long prev = mergeSort(c1);
		int j = 0;
		int sol = 0;
		while ((1<<j) <= max) {
			for (int i = 0; i < n; i++) c1[i] = a[i]^sol^(1<<j);
			long ans1 = mergeSort(c1);
			if (ans1 < prev) {
				sol^=(1<<j);
				prev = ans1;
			}
			j++;
		}
		
		out.println(prev + " " +sol);
		
		
		out.close();
	}

	
	static long mergeSort(int[] a) {
		if (a.length == 1) return 0;
		int[] l = new int[a.length/2];
		int[] r = new int[a.length-a.length/2];
		for (int i = 0; i < l.length; i++) l[i]=a[i];
		for (int i = 0; i < r.length; i++) r[i] = a[i+l.length];
		return mergeSort(l) + mergeSort(r) + merge(l,r,a);
	}
	
	static long merge(int[] a, int[] b, int[] c) {
		// Count inversions too
		int i = 0, j = 0, k = 0;
		long count = 0;
		while (i < a.length && j < b.length && k < c.length) {
			if (a[i] <= b[j]) {
				// Expected.
				c[k]=a[i];
				i++;
			}
			else if (a[i] > b[j]) {
				// Incr.
				c[k]=b[j];
				count += a.length-i;
				j++;
			}
			k++;
		}
		while (i < a.length) {
			c[k]=a[i];
			i++;
			k++;
		}
		while (j < b.length) {
			c[k]=b[j];
			j++;
			k++;
		}
		return count;
		
	}
	
}
