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.Applicative
import           Control.Monad
import qualified Data.ByteString.Char8 as B

import           Data.Array.IArray

solve :: IO ()
solve = do
    xs <- B.getLine
    ys <- B.getLine
    print $ lcs xs ys

lcs :: B.ByteString -> B.ByteString -> Int
lcs xs ys = go (lx, ly)
  where
    lx = B.length xs - 1
    ly = B.length ys - 1

    m :: Array (Int, Int) Int
    m = listArray ((-1, -1), (lx, ly))
      $ fmap go [(i, j) | i <- [-1 .. lx], j <- [-1 .. ly]]

    go (i, j)
        | i < 0 || j < 0 = 0
        | x == y         = 1 + m ! (i - 1, j - 1)
        | x /= y         = max (m ! (i, j - 1)) (m ! (i - 1, j))
      where
        x = xs `B.index` i
        y = ys `B.index` j

main :: IO ()
main = do
    n <- readLn
    replicateM_ n solve