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)

import Control.Monad (replicateM_)
import Data.Array (listArray, (!))

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

lcs :: (String, String) -> Int
lcs (xs,ys) = lcs' lx ly
    where
        lx = length xs
        ly = length ys
        xs' = listArray (0,lx-1) xs
        ys' = listArray (0,ly-1) ys
        lcs' 0 _ = 0
        lcs' _ 0 = 0
        lcs' i j | xs' ! (i-1) == ys' ! (j-1) = lcs' (i-1) (j-1) + 1
                 | otherwise = lcs' i (j-1) `max` lcs' (i-1) j