Problem:
0-1 Knapsack Problem

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 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 ≤ 1000

1 ≤ wi ≤ 1000

1 ≤ W ≤ 10000

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)

{-# LANGUAGE FlexibleContexts #-}

import Control.Applicative
import Data.Array.Unboxed

main = do
  [_,capa] <- map (read :: String -> Int) . words <$> getLine
  goods <- map (map (read :: String -> Int) . words) . lines <$> getContents
  print $ maximum $ elems (resolve capa (listArray (0,capa) [0,0..]) goods)

resolve :: Int -> Array Int Int -> [[Int]] -> Array Int Int
resolve capa capaArr ([v,w]:xs)
  | xs == [] = koushin capaArr [v,w] (capa-w)
  | otherwise = resolve capa (koushin capaArr [v,w] (capa-w)) xs
  where
    koushin :: Array Int Int -> [Int] -> Int -> Array Int Int
    koushin capaArr [v,w] n
      | n == 0 && capaArr ! w <= v = capaArr // [( w , v)]
      | n == 0 = capaArr
      | capaArr ! n == 0       = koushin capaArr [v,w] (n-1)
      | capaArr ! (n+w) >= v + (capaArr ! n)  = koushin capaArr [v,w] (n-1)
      | otherwise = koushin (capaArr // [( n + w , v + (capaArr ! n))]) [v,w] (n-1)