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.*;
import java.math.*;
import java.io.*;
import java.text.*;

public class cf533 {
	
	static Scanner scan = new Scanner(System.in);
	static int MAX = 1111111; 
	public static void main(String[] args) {
		int N = scan.nextInt(); 
		int arr[] = new int [N]; for(int i=0; i<N; i++) {arr[i] = scan.nextInt();}
		
		int answer = Integer.MAX_VALUE , buffer = 0; 
		for(int i=1 ;i<111; i++) {
			int temp =  solve(0, MAX , arr , i); 
			if(temp < answer) {answer = temp; buffer = i;}
		}
		System.out.println(buffer +" "+ answer);
		
	}
	
	public static int solve(int left, int right, int[] arr, int t) {
		if(left > right) {return -1; }
		else if(left == right) {return left; }
		else if(right-left == 1) {
			if(check(arr,t,left)) {return left;}
			else {return right;}
		}else {
			int mid = (left+right)/2; 
			if(check(arr,t,mid)) {return solve(left, mid, arr,t);}
			else {return solve(mid+1,right,arr,t);}
		}
	}
	
	public static boolean check(int[] arr, int t, int cost) {
		
		for(int i=0 ;i<arr.length; i++) {
			int minCurrcost = Integer.min(Math.abs(arr[i]-t-1), Math.abs(arr[i]-t));
			minCurrcost = Integer.min(minCurrcost, Math.abs(arr[i]-t+1)); 
			if(minCurrcost > cost) {return false;}
			else {cost -= minCurrcost;}
		}
		return true; 
	}

}
