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: Time Limit Exceeded)

lcs :: (String,String) -> Int -> Int
lcs ([],_) n = n
lcs (_,[]) n = n
lcs (ax@(x:xs),ay@(y:ys)) n
  | x == y    = lcs (xs,ys) (n+1)
  | otherwise = max (lcs (ax,ys) n) (lcs (xs,ay) n)
 
main = do
  getLine
  inputs <- fmap lines getContents
  let targets = zip (takeOnlyOdd inputs) (takeOnlyEven inputs)
  mapM_ putStrLn $ map (\x -> show $ lcs x 0) targets
 
takeOnlyEven :: [a] -> [a]
takeOnlyEven xs = map snd . filter (\(i, n) -> i `mod` 2 == 0) $ zip [1..] xs
takeOnlyOdd :: [a] -> [a]
takeOnlyOdd xs = map snd . filter (\(i, n) -> i `mod` 2 == 1) $ zip [1..] xs