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: Accepted)

{-# LANGUAGE BangPatterns #-}
import Control.Monad
import Data.Array.Unboxed
import Data.Functor
import Data.List
import qualified Data.ByteString.Char8  as B
import qualified Data.IntMap.Strict     as IntMap
import qualified Data.Vector.Unboxed    as VU

main :: IO ()
main = do
  n <- readLn
  xs <- replicateM n get2Lines :: IO [(String, String)]
  mapM_ (print . uncurry rowdp') xs 

get2Lines :: IO (String, String)
get2Lines = do
  a <- getLine
  b <- getLine
  return (a, b)

rowdp' :: String -> String -> Int
rowdp' xs ys = last $ foldl' scanY headRow xs
  where
    m = length ys
    headRow = replicate (m + 1) 0 :: [Int]
    scanY :: [Int] -> Char -> [Int]
    scanY prevRow x = myscanl' f 0 yvprev
      where
        yvprev = zip3 ys prevRow $ tail prevRow
        f :: Int -> (Char, Int, Int) -> Int
        f v (y, a, b) = if x == y 
                        then 1 + a
                        else max b v

myscanl'           :: (b -> a -> b) -> b -> [a] -> [b]
myscanl' = scanlGo'
  where
    scanlGo'           :: (b -> a -> b) -> b -> [a] -> [b]
    scanlGo' f !q ls    = q : (case ls of
                            []   -> []
                            x:xs -> scanlGo' f (f q x) xs)