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: Time Limit Exceeded)

{-# LANGUAGE BangPatterns #-}
module Main where
import qualified Data.HashSet as HS
import qualified Data.Vector.Mutable as MV
import qualified Data.Vector as V
import Control.Monad

readL :: Read a => IO [a]
readL = map read . words <$> getLine

main = do
  [n, duration] <- readL
  jobs <- replicateM n $ (\[a,b] -> (a,b)) <$> readL

  result <- MV.replicate (duration+1) (0 :: Int, HS.empty :: HS.HashSet Int)
  forM_ [0..duration] $ \day -> do
    current <- V.freeze result
    forM_ (zip [0..] jobs) $ \(jobId, (delay, income)) -> do
      let day' = day + delay
      unless (day' > duration) $ do
        forM_ [day..duration] $ \baseDay -> do
          let (total, done) = current V.! baseDay
          unless (HS.member jobId done) $ do
            let !new = total + income
            let !done' = HS.insert jobId done
            (old, _) <- MV.read result day'
            when (new > old) $ do
              MV.write result day' (new, done')

  V.unsafeFreeze result >>= print . V.maximum . V.map fst