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.*;
import java.lang.*;
import java.rmi.*;
import java.util.*;
import java.math.*;

public class Main {

    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

    static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    static double nextDouble() throws IOException {
        in.nextToken();
        return in.nval;
    }

    static long nextLong() throws IOException {
        in.nextToken();
        return (long) in.nval;
    }

    static String next() throws IOException {
        in.nextToken();
        return in.sval;
    }

    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static int a[];

    public static void main(String[] args) throws IOException {
        int n, m;
        n = nextInt();
        m = nextInt();
        int k = m;
        a = new int[n + 10];
        int _max = 0;
        for (int i = 1; i <= n; ++i) {
            a[i] = nextInt();
            _max = Math.max(_max, a[i]);
        }
        for (int i = 1; i <= n; ++i) {
            m -= (_max - a[i]);
        }
        if (m <= 0) {
            out.print(_max);
        } else {
            if (m % n != 0) {
                out.print(_max + m / n + 1);
            } else {
                out.print(_max + m / n);
            }
        }
        out.print(' ');
        out.println(_max + k);
        out.flush();
    }
}