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 math

n = int(input().strip())
m = int(input().strip())
init_people = []
for _ in range(n):
    init_people.append(int(input().strip()))

init_people.sort()
# i = 1
# max_k = init_people[-1] + m
# if n == 1:
#     print(max_k, max_k)
# else:
#     while m > 0:
#         if i == n:
#             init_people.sort()
#             j = 0
#             while m > 0:
#                 init_people[j % n] += 1
#                 m -= 1
#                 j += 1
#             break
#         j = 0
#         while j < i:
#             if init_people[j] < init_people[i]:
#                 init_people[j] += 1
#                 j += 1
#                 m -= 1
#             else:
#                 i += 1
#                 break
#
#     print(min(init_people), max_k)



min_k = init_people[0]
i = 1
m_min = m
while m_min > 0:
    if i == n:
        min_k += math.ceil(m_min / n)
        break

    diff = init_people[i] - init_people[i - 1]
    sub = min(diff, math.ceil(m_min / i))
    m_min -= i * sub
    min_k += sub
    i += 1
if n == 1:
    min_k = init_people[0] + m
# elif n == 2:
#     m_min = m - init_people[1] + init_people[0]
#     min_k = init_people[1] + math.ceil(m_min / 2)

print(max(min_k, max(init_people)), init_people[-1] + m)

 	   	 		   	 				 	  	  	 	 		