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.
//package codeforcesa4;
import java.util.*;
public class Benches {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int m=in.nextInt();
        int a[]=new int[n];
        int ansmin=0;
        int ansmax=0;
        int max=0;
        for(int i=0;i<n;i++)
        {
            a[i]=in.nextInt();
            if(a[i]>max){
                max=a[i];
            }
        }
        ansmax=max+m;
        Arrays.sort(a);
        int x=0;
        for(int i=a.length-2;i>=0;i--)
        {
            //System.out.println("a[i]="+a[i]);
            x=(max-a[i]);
            //System.out.println("x="+x);
            if(x<=m){
                a[i]=a[i]+x;
                m=m-x;
                }
            else{
                a[i]=a[i]+m;
                m=0;break;
                }
        }
        //System.out.println("m="+m);
        if(m==0){
            int min=0;
            for(int i=0;i<n;i++)
            {
                if(a[i]>min){
                    min=a[i];
                }
            }
            ansmin=min;
        }
        else{
            if(m%n==0)
                ansmin=a[0]+(m/n);
            else
                ansmin=a[0]+(m/n)+1;
            
        }
        System.out.println(ansmin+" "+ansmax);
        
    }
}
