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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author math2do
*/public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
static class TaskA   {
public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        int[] arr = new int[n];
        int maxp = Integer.MIN_VALUE;
        int minp = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
            maxp = Math.max(maxp, arr[i]);
            minp = Math.max(minp, arr[i]);
        }
        maxp += m;
        Boolean found = false;
        for (int i = 0; i < n; i++) {
            int red = minp - arr[i];
            if (red <= m) {
                m -= red;
            } else {
                found = true;
                break;
            }
        }
        if (!found) {
            minp += m / n + (m % n != 0 ? 1 : 0);
        }
        out.print(minp + " " + maxp);
    }

}
static class InputReader   {
public BufferedReader reader;
public StringTokenizer tokenizer;

public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

}
}

