Salem gave you n sticks with integer positive lengths a_1, a_2, …, a_n.

For every stick, you can change its length to any other positive integer length (that is, either shrink or stretch it). The cost of changing the stick's length from a to b is |a - b|, where |x| means the absolute value of x.

A stick length a_i is called almost good for some integer t if |a_i - t| ≤ 1.

Salem asks you to change the lengths of some sticks (possibly all or none), such that all sticks' lengths are almost good for some positive integer t and the total cost of changing is minimum possible. The value of t is not fixed in advance and you can choose it as any positive integer. 

As an answer, print the value of t and the minimum cost. If there are multiple optimal choices for t, print any of them.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of sticks.

The second line contains n integers a_i (1 ≤ a_i ≤ 100) — the lengths of the sticks.

Output

Print the value of t and the minimum possible cost. If there are multiple optimal choices for t, print any of them.

Examples

Input


3
10 1 4


Output


3 7


Input


5
1 1 2 2 3


Output


2 0

Note

In the first example, we can change 1 into 2 and 10 into 4 with cost |1 - 2| + |10 - 4| = 1 + 6 = 7 and the resulting lengths [2, 4, 4] are almost good for t = 3.

In the second example, the sticks lengths are already almost good for t = 2, so we don't have to do anything.

import java.util.Arrays;
import java.util.Scanner;

public class salem_sticks1105A {
public static void main(String[]   args) {
	Scanner sc=new Scanner (System.in);
	int n=sc.nextInt();
	int arr[]=new int[n];
	for(int i=0;i<n;i++)
		arr[i]=sc.nextInt();
	Arrays.sort(arr);
	int min=arr[0];
	int max=arr[n-1];
	int t=min;
	int cost=50000000;
	int c=0;
	int v=0;
	for(int i=min;i<=max;i++) {
		c=0;
		for(int j=0;j<n;j++) {
		if(arr[j]>i) {
			v=Math.abs(arr[j]-(i+1));
			if(v>0)
			  c+=Math.abs(arr[j]-(i+1));
		}
		else {
			if(arr[j]<i) {
				 v=Math.abs((i-1)-arr[j ]);
				 
				 if(v>0)
				c+=Math.abs((i-1)-arr[j]);
			}
		}
		}
		if(c<cost) {
			cost=c;
			t=i;
		
	}
	}
	System.out.println(t+" "+cost);
	
	
	
	
	
}
}
