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 sys

def function(data, m):
	"""
	input:
		data: a list of integer
		m: value to split
	"""
	if m == 0: return data
	if len(data) == 1:
		data[0] += m
		return data

	min_value = min(data)
	max_value = max(data)
	if max_value - min_value >= m:
		return data
	else:
		if len(set(data)) == 1:
			for i in range(len(data)):
				data[i] += 1
				m -= 1
		else:
			for i in range(len(data)):
				if data[i] < max_value:
					data[i] += 1
					m -= 1
		return function(data, m)

data = []
T = 0
for line in sys.stdin:
	value = int(line)
	if T == 0:
		T = value
	data.append(value)
	if T + 2 == len(data): break

# data = [4, 6, 1, 1, 1, 1]
# data = [1, 10, 5]
# data = [3, 6, 1, 6, 5]
# data = [3, 7, 1, 6, 5]

n, m = data[0:2]
max_possible_k = max(data[2:]) + m
min_possible_k = max(function(data[2:], m))
print min_possible_k, max_possible_k