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.
all = input().split()[0]
all = int(all)
man = input().split()[0]
man = int(man)
min = 0
max = 0
yz = []

for i in range(all):
    yz.append(int(input().split()[0]))
yz.sort()
max = yz[-1] + man
tmp = 0
if all == 1:
    min = max
    print(min, max)
    exit(0)
# print(yz , min)03.py:13
while True:
    for i in range(all):
        if (i == all-1)&(yz[i]==yz[i-1]-1):
            yz[i] = yz[i] + 1
            tmp += 1
        if (i == all-1)&(yz[i]!=yz[i-1]-1):
            break
        if (yz[i] == yz[i+1]) & (tmp != man):
            yz[i] = yz[i]+1
            tmp += 1
        elif (yz[i] < yz[i+1]) & (tmp != man):
            yz[i] = yz[i] + 1
            tmp += 1
            break
        else:
            break
    if tmp == man:
        break
min = sorted(yz)[-1]
print(min , max)
		 			  	 			   				  	 	     	