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.Monad (replicateM_)
import qualified Data.Vector as V

main :: IO ()
main = do
    n <- readLn
    replicateM_ n $ do
        str1 <- getLine
        str2 <- getLine
        print $ lcs (str1,str2)


lcs :: (String,String) -> Int
lcs (xs,ys) = lcsTable (lx,ly)
    where
        lx = length xs
        ly = length ys
        xs' = V.fromList xs
        ys' = V.fromList ys
        idx (x,y) = x + y*(lx+1)
        xdi i = (i `mod` (lx+1), i `div` (lx+1))
        lcsTable = (V.map (lcs'.xdi) (V.enumFromN 0 (idx (lx,ly)+1)) V.!) . idx
        lcs' (0,_) = 0
        lcs' (_,0) = 0
        lcs' (i,j) 
          | xs' V.! (i-1) == ys' V.! (j-1) = lcsTable (i-1,j-1) + 1
          | otherwise = lcsTable (i,j-1) `max` lcsTable (i-1,j)