Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Data.List
import Control.Monad
import qualified Data.IntMap  as IM

main = do
  [n,m] <- map read.words<$>getLine
  ab <- replicateM n $ (\[a,b]->(read a,read b)).words<$>getLine
  print $ solve n m ab

--solve :: Int -> Int -> [(Int,Int)] -> Int
solve n m ab = go IM.empty 0 abs where
  go set _ [] = sum $ map (uncurry (*)) $ IM.toList set
  go set n (ab:abs) = go next newn abs where
    newn = m-fst(head ab)+1
    next = foldl inner (IM.insert 0 (newn-n) set) $ f ab
    inner set (n,b) = insertIfLarge b set
  abs = groupBy (\(x,_)(y,_)->x==y)
      $ sortBy (\(x,_)(y,_)->compare y x)
      $ sortBy (\(_,x)(_,y)->compare y x) ab
  f a@((n,_):_) = take (m-n+1) a

insertIfLarge a set = case lookupMin set of
  Nothing -> set
  Just (k, 1) -> if a > k then IM.insertWith(+) a 1 $ IM.delete k set else set
  Just (k, _) -> if a > k then IM.insertWith(+) a 1 $ IM.adjust(subtract 1) k set else set

lookupMin im
  | IM.null im = Nothing
  | otherwise = Just $ IM.findMin im