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 Ruby (judge verdict: Wrong Answer)

def catch(a,b,dp,x,y)
  if dp[x-1][y]<dp[x][y-1] then
    dp[x][y]=dp[x][y-1]
  elsif dp[x-1][y]>dp[x][y-1]
    dp[x][y]=dp[x-1][y]
  else
    dp[x][y]=dp[x-1][y]
    if a==b then
      dp[x][y]+=1
    end
  end
end



n=gets.to_i
arr=Array.new
n.times do |i|
  a=gets.chomp!.split("")
  b=gets.chomp!.split("")
  dp=Array.new(a.length+1){Array.new(b.length+1,0)}
  1.upto(a.length) do |x|
     1.upto(b.length) do |y|
      catch(a[x-1],b[y-1],dp,x,y)
    end
  end
  arr << dp[-1][-1]
  a.clear
  b.clear
end
for x in arr do
  print x,"\n"
end