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.Scanner;

public class benches {

    public int[] find_num(int[] a,int m)
    {
        int n;
        n=a.length;
        int[] k;
        k=new int[2];

        this.sort(a,n);

        k[0]=a[n-1]+m;
        k[1]=k_min(a,m);

        return(k);

    }

public static void sort(int[] a,int n) {
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - i - 1; j++) {
            if (a[j] > a[j + 1]) {
                int temp = a[j];
                a[j] = a[j + 1];
                a[j + 1] = temp;

            }
        }
    }
}


public int k_min(int[] a,int m)
{
    int x,y;
    y=0;
    int i,n;
    n= a.length ;
    for(i=0;i<n-1;i++)
    {
        m = m - (a[n - 1] - a[i]);
        if (m <= 0)
            return (a[n - 1]);
    }

    x= m/n;
    if(m%n!=0)
    y=1;
    return(a[n-1]+x+y);
}


    public static void main(String[] args) {
        int[] a ;
        int[] k= new int[2];
        int m,n;
        benches b1 = new benches();
        Scanner sc= new Scanner(System.in);
        n=sc.nextInt();
        a= new int[n];
       m= sc.nextInt();

        for(int i=0;i<n;i++)
        {
            a[i]=sc.nextInt();
        }

     sort(a,n);
 //       for(int i=0;i<n;i++)
 //           System.out.print(a[i]);

        k= b1.find_num(a,m);
       System.out.print(k[1]+"  "+k[0]);
    }



}
