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 Rating{
	public static Scanner in=new Scanner(System.in);
	public static long ans[][]=new long[31][2];
	public static int a[]=new int[300005],b[]=new int[300005],n;
	public static void dfs(int l,int r,int d) {
		if(d==-1)return;
		int pos1=l,pos2=l;
		for(int i=l;i<=r;++i)if((a[i]>>d&1)==1) {
			b[pos2++]=a[i];
			ans[d][1]+=pos1-l;
		}
		else {
			a[pos1++]=a[i];
			ans[d][0]+=pos2-l;
		}
		for(int i=pos1;i<=r;++i)a[i]=b[i-pos1+l]; 
		if(l!=pos1)dfs(l,pos1-1,d-1);if(pos2!=l)dfs(pos1,r,d-1);
	}
	public static void main (String[] args) throws java.lang.Exception{
		n=in.nextInt();
		long res=0,x=0;
		for(int i=1;i<=n;++i)a[i]=in.nextInt();
		dfs(1,n,30);
		for(int i=30;i>=0;--i)if(ans[i][0]<=ans[i][1])res+=ans[i][0];else {res+=ans[i][1];x|=(1<<i);}
		System.out.println(res+" "+x);
	}
}