Problem:
Score: 200 points

Problem Statement

The restaurant AtCoder serves the following five dishes:

ABC Don (rice bowl): takes A minutes to serve.

ARC Curry: takes B minutes to serve.

AGC Pasta: takes C minutes to serve.

APC Ramen: takes D minutes to serve.

ATC Hanbagu (hamburger patty): takes E minutes to serve.

Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.

This restaurant has the following rules on orders:

An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).

Only one dish can be ordered at a time.

No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.

E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.

Here, he can order the dishes in any order he likes, and he can place an order already at time 0.

Constraints

A, B, C, D and E are integers between 1 and 123 (inclusive).

Input

Input is given from Standard Input in the following format:

A
B
C
D
E

Output

Print the earliest possible time for the last dish to be delivered, as an integer.

Sample Input 1

29
20
7
35
120

Sample Output 1

215

If we decide to order the dishes in the order ABC Don, ARC Curry, AGC Pasta, ATC Hanbagu, APC Ramen, the earliest possible time for each order is as follows:

Order ABC Don at time 0, which will be delivered at time 29.

Order ARC Curry at time 30, which will be delivered at time 50.

Order AGC Pasta at time 50, which will be delivered at time 57.

Order ATC Hanbagu at time 60, which will be delivered at time 180.

Order APC Ramen at time 180, which will be delivered at time 215.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 2

101
86
119
108
57

Sample Output 2

481

If we decide to order the dishes in the order AGC Pasta, ARC Curry, ATC Hanbagu, APC Ramen, ABC Don, the earliest possible time for each order is as follows:

Order AGC Pasta at time 0, which will be delivered at time 119.

Order ARC Curry at time 120, which will be delivered at time 206.

Order ATC Hanbagu at time 210, which will be delivered at time 267.

Order APC Ramen at time 270, which will be delivered at time 378.

Order ABC Don at time 380, which will be delivered at time 481.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 3

123
123
123
123
123

Sample Output 3

643

This is the largest valid case.

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

main :: IO()
main = do
    a <- readLn
    b <- readLn
    c <- readLn
    d <- readLn
    e <- readLn
    print $ solve [a,b,c,d,e]



solve :: [Int] -> Int
solve ls = sum summing + last changeLs
    where changeLs = firstSmall ls []
          summing = map floorUp (init changeLs)

--1の位が最も小さい数字を一番後ろにに持っていく。ただし０は１０と扱う。
firstSmall :: [Int] -> [Int] -> [Int]
firstSmall ls resultLs
    | ls == []							= resultLs
    | resultLs == []					= firstSmall (tail ls) [head ls]
    | otherwise 						= if (getOne $ head ls) > (getOne $ last resultLs)
												then firstSmall (tail ls) ((head ls):resultLs)
												else firstSmall (tail ls) (resultLs ++ [(head ls)])

--１の位を出力。ただし0だったら10を返す。
getOne :: Int -> Int
getOne num = if (mod num 10) == 0 then 10 else mod num 10 

--１０の位までで切り上げ。
floorUp :: Int -> Int
floorUp num = if mod num 10 == 0 then num else num + 10 - (mod num 10)