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.*;
import java.io.*;
 
public class Prog {
  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 max = -200;
    int sum = m;
    for (int i = 0; i < n; i++) {
      a[i] = in.nextInt(); 
      if (a[i] > max) max = a[i];
      sum += a[i];
    }
    Arrays.sort(a);
    int[] b = a.clone();
    int l = -1;
    int r = sum	+ 1;
    while (l < r - 1) {
      a = b.clone();
      int k = (l + r) / 2;
      if (bs(k, n, a, m)) 
        l = k;                                                            
      else
        r = k;
    } 
    System.out.println(l + " " + (max + m));
  }
  public static boolean bs(int k, int n, int[] a, int m) {
    while (m > 0) {
      a[0]++;
      m--;
      Arrays.sort(a);
    }
    int max = -200;
    for (int i = 0; i < n; i++) {
      if (a[i] > max) max = a[i];
    }
    return (k <= max);
  }
}