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.*;

public class Demo
{
       public static void main(String args[])
       {
              Scanner sc=new Scanner(System.in);
              int n=sc.nextInt();
              int a[]=new int[n];
              float sum=0;
              for(int i=0;i<n;i++)
              {
                     a[i]=sc.nextInt();
                     sum=sum+a[i];
              }
              /*double ans=(sum/n);
              //System.out.print(ans);
              ans=Math.ceil(ans);
              
              //System.out.print(ans);
              int t=0;
              for(int i=0;i<n;i++)
              {
                     if(Math.abs(ans-a[i])>1)
                     {
                            t=t+((int)Math.abs(ans-a[i]))-1;
                     }
              }*/
              int min=100000000;
              int temp=min;
              int ans=0;
              for(int i=1;i<=100;i++)
              {
                     int t=0;
                     for(int j=0;j<n;j++)
                     {
                            if(Math.abs(i-a[j])>1)
                            {
                                   t=t+((int)Math.abs(i-a[j]))-1;
                            }
                     }
                     min=Math.min(min,t);
                     if(temp!=min)
                     {
                            ans=i;
                            temp=min;
                     }
              }
              //int ans1=(int)ans;
              System.out.print(ans+" "+min);
       }
}