Problem:
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Longest Common Subsequence

For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \{a,b,c,b,d,a,b\}$ and $Y = \{b,d,c,a,b,a\}$, the sequence $\{b,c,a\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\{b,c,a\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\{b,c,b,a\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\{b,c,b,a\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.

Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Input

The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.

Output

For each dataset, print the length of LCS of $X$ and $Y$ in a line.

Constraints

$1 \leq q \leq 150$

$1 \leq$ length of $X$ and $Y$ $\leq 1,000$

$q \leq 20$ if the dataset includes a sequence whose length is more than 100

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Haskell (judge verdict: Memory Limit Exceeded)

import           Control.Applicative
import           Control.Monad
import qualified Data.ByteString.Char8 as BS
import qualified Data.List             as L
import qualified Data.Map              as M
import           Data.Maybe
import qualified Data.Set              as S
import qualified Data.Tuple            as T


readInt' :: BS.ByteString -> Int
readInt' = fst . fromJust . BS.readInt

-- lcs :: BS.ByteString
--     -> BS.ByteString
--     -> Int
lcs :: (Ord a, Num a) => BS.ByteString -> BS.ByteString -> M.Map (Int, Int) a
lcs str1 str2 =
    L.foldl' (\ acc (i, j) ->
                  if BS.index str1 (i - 1) == BS.index str2 (j - 1)
                  then M.insert (i, j)
                           (M.findWithDefault 0 (i - 1, j - 1) acc + 1)
                           acc
                  else M.insert (i, j)
                           (max (M.findWithDefault 0 (i - 1, j) acc)
                                (M.findWithDefault 0 (i, j - 1) acc))
                           acc)
         (M.singleton (0, 0) 0)
         [(i, j) | i <- [1..BS.length str1], j <- [1..BS.length str2]]


main :: IO ()
main = do
  n <- readLn :: IO Int
  forM_ [1..n] $ \_ -> do
         str1 <- BS.getLine
         str2 <- BS.getLine
         let dp = lcs str1 str2
         putStrLn (show $ M.findWithDefault 0 (BS.length str1, BS.length str2)
                        dp)