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 #-}

solve :: (String, String) -> Int
solve (xss, yss) = lcs xss yss befMemo [0]
  where
    lcs [] _ memo _     = last memo
    lcs (x:xs) [] _  as = lcs xs yss (reverse as) [0]
    lcs xs ys (b:bs) as = lcs xs (tail ys) bs (m : as)
      where
        !m = if (head xs) == (head ys) then b + 1 else max (head as) (head bs)
    befMemo = replicate (length yss + 1) 0

main :: IO ()
main =  getLine
        >>  fmap lines getContents
        >>= toPair []
        >>= mapM_ (print . solve)
  where
    toPair pairs [] = return $ reverse pairs
    toPair pairs (xs:ys:rest) = toPair ((xs, ys) : pairs) rest