Problem:
0-1 Knapsack Problem II

You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.

The items have combined weight at most W, that is capacity of the knapsack.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1
v2 w2
:
vN wN

The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100

1 ≤ vi ≤ 100

1 ≤ wi ≤ 10,000,000

1 ≤ W ≤ 1,000,000,000

Sample Input 1

4 5
4 2
5 2
2 1
8 3

Sample Output 1

13

Sample Input 2

2 20
5 9
4 10

Sample Output 2

9

Solution in Haskell (judge verdict: Memory Limit Exceeded)

{-# OPTIONS_GHC -O2 #-}
import Control.Applicative
import Data.IORef
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as V
import Control.Monad
 
main = do
  [_,capa] <- map (read :: String -> Int) . words <$> getLine
  goods <- map (map (read :: String -> Int) . words) . lines <$> getContents
  capaVM <- VM.replicate (capa+1) 0
  ans <- newIORef 0
  mapM_ (resolve capa capaVM ans) goods
  print =<< readIORef ans
 
resolve capa capaVM ans [v,w] =
  if capa < w
    then return ()
    else V.mapM_ (koushin [v,w]) (V.enumFromStepN (capa-w) (-1) (capa-w+1))
       where
         koushin [v,w] n = do
           oldValue <- VM.read capaVM n
           newValue <- VM.read capaVM (n+w)
           judge oldValue newValue
             where
               judge oldValue newValue
                 | n == 0 && newValue <= v  = kakikae
                 | n ==0                    = return ()
                 | newValue >= v + oldValue = return ()
                 | otherwise  = kakikae
                 where
                   kakikae = do
                     VM.write capaVM (n+w) (v + oldValue)
                     ans' <- readIORef ans
                     writeIORef ans (max ans' (v + oldValue))