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.
import java.util.Arrays;
import java.util.Scanner;
public class Main{
	
	final static int MAX_SIZE=105;
	static int[] Benches=new int[MAX_SIZE];
	
	public static void main(String[] agrs) {
		
		Scanner in=new Scanner(System.in);
		int n,m;
		int Max_Case=0;
		int Sum=0;
		n=in.nextInt();
		m=in.nextInt();
		for(int i=1;i<=n;i++) {
			Benches[i]=in.nextInt();
			if(Benches[i]>Max_Case) {
				Max_Case=Benches[i];
			}
			Sum+=Benches[i];
		}
		int Min_Res=0;
		int Aver=Max_Case*n;
		if(Aver-Sum<m) {
			Min_Res=(m-Aver+Sum)/n+Max_Case;
			if((m-Aver+Sum)%n!=0) {
				Min_Res++;
			}
		}else {
			Min_Res=Max_Case;
		}
		
		int Max_Res=m+Max_Case;
		
		System.out.println(Min_Res+" "+Max_Res);
		
	}
}
