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 whatever //do not write package name here */

import java.io.*;
import java.util.*;
import java.lang.*;

public class GFG {
	public static void main (String[] args) {
		//System.out.println("GfG!");
		
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		
		int[] a=new int[n];
		for(int i=0;i<n;i++)
		{
		    int x=sc.nextInt();
		    a[i]=x;
		}
		Arrays.sort(a);
		int max=a[n-1]+m,min=a[n-1];
		int ctr=1;
		for(int i=0;i<n-1;i++)
		{
		    /*m=m-(a[n-1]-a[i]);
		    if(m<=0)
		    {
		        min=a[n-1];
		        break;
		    }*/
		    if(a[i+1]>a[i])
		    {
		        //m=m-(a[i+1]-a[i])*ctr;
		        for(int j=0;j<=i;j++)
		        {
		            m=m-(a[i+1]-a[j]);
		            a[j]=a[i+1];
		            //System.out.println(m);
		        }
		        ctr=1;
		    }
		    else
		    {
		        ctr++;
		    }
		    if(m<=0)
		    {
		        min=a[n-1];
		        break;
		    }
		}
		if(m>0)
		{
		    if(n>m)
		    {
		        min=a[0]+1;
		    }
		    else
		    {
		        min=a[0]+m/n;
		        if(m%n>0)
		        min++;
		    }
		}
		
		System.out.println(min+" "+max);
	}
}