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)

{-# LANGUAGE BangPatterns #-}
import Control.Monad
import qualified Data.ByteString.Char8  as B
import Data.Array.Unboxed
import Data.Functor

main :: IO ()
main = do
  n <- readLn
  replicateM n iterateSolveDp
  return ()

iterateSolveDp = do
  x <- B.unpack <$> B.getLine
  y <- B.unpack <$> B.getLine
  let n = length x
      m = length y
      a = listArray (0, n-1) x :: Array Int Char
      b = listArray (0, m-1) y :: Array Int Char
      dp :: Array (Int, Int) Int
      dp = listArray ((0, 0), (n, m)) $ map f [(i, j)|i<-[0..n],j<-[0..m]]
      f (0, _) = 0
      f (_, 0) = 0
      f (i, j)
        | a!(i-1) == b!(j-1) = 1 + dp!(i-1, j-1)
        | otherwise = max (dp!(i, j-1)) (dp!(i-1, j))
  print $ dp!(n, m)