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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;
import java.util.SortedSet;
import java.util.TreeSet;
public class Benches {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader x = new BufferedReader(new InputStreamReader(System.in));
		 int n=Integer.parseInt(x.readLine());
		 int m=Integer.parseInt(x.readLine());		 
		 Integer[] y=new Integer[n];
		 for(int i=0; i<n; i++) {			
			 y[i]=Integer.parseInt(x.readLine());
		 }		
		 Arrays.sort(y);		
		 SortedSet<Integer> b2=new TreeSet<>();	
		 int min=y[y.length-1];
			 int max=min+m;
			
			 for(int i=0;i<m;i++){			
				 y[0]=y[0]+1;
				 Arrays.sort(y);			
				 b2.addAll(Arrays.asList(y));				
			 }		 
		 System.out.println(((m<min-y[0])?min:b2.last())+" "+max) ;
		 }
}
