There are n benches in the Berland Central park. It is known that a_i people are currently sitting on the i-th bench. Another m people are coming to the park and each of them is going to have a seat on some bench out of n available.

Let k be the maximum number of people sitting on one bench after additional m people came to the park. Calculate the minimum possible k and the maximum possible k.

Nobody leaves the taken seat during the whole process.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of benches in the park.

The second line contains a single integer m (1 ≤ m ≤ 10 000) — the number of people additionally coming to the park.

Each of the next n lines contains a single integer a_i (1 ≤ a_i ≤ 100) — the initial number of people on the i-th bench.

Output

Print the minimum possible k and the maximum possible k, where k is the maximum number of people sitting on one bench after additional m people came to the park.

Examples

Input

4
6
1
1
1
1


Output

3 7


Input

1
10
5


Output

15 15


Input

3
6
1
6
5


Output

6 12


Input

3
7
1
6
5


Output

7 13

Note

In the first example, each of four benches is occupied by a single person. The minimum k is 3. For example, it is possible to achieve if two newcomers occupy the first bench, one occupies the second bench, one occupies the third bench, and two remaining — the fourth bench. The maximum k is 7. That requires all six new people to occupy the same bench.

The second example has its minimum k equal to 15 and maximum k equal to 15, as there is just a single bench in the park and all 10 people will occupy it.

/*	Basic import definitions	*/
	import java.io.*;
	import java.util.*;
	import java.math.BigInteger;
	import java.util.LinkedList;
	import java.io.BufferedReader;
	import java.io.IOException;
	import java.io.InputStreamReader;

/*	Basic Class definitions*/
	public class A1042
	{
		public static void main(String[] args) 
		{
			Scanner sc = new Scanner(System.in);

			int i , j , k;
            int n,m;
            long sum = 0;
            int max = 0;
            int ans;

            n = sc.nextInt();
            m = sc.nextInt();

			for ( i = 0 ; i < n ; i++ ) 
			{
                k = sc.nextInt();
                max = (int)Math.max(k, max);
                sum += k;
            }
            
            long cat = sum + m - max*n;
            if(cat < 0)
            {
                ans = max;
            }
            else
            {
                ans = max+(int)Math.ceil((double)cat/n);
            }
            System.out.println(ans+" "+(max+m));
			sc.close();
		}
	}