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: Time Limit Exceeded)

class SequencePair

  def initialize(seq1, seq2)
    @seq1 = seq1.chomp
    @seq2 = seq2.chomp
    @seq1_len = @seq1.length
    @seq2_len = @seq2.length
    @lcs_table = Array.new(seq1.length).map{Array.new(seq2.length,0)}
  end

  def lcs_len
    if @seq1_len*@seq2_len==0
      return 0
    else
      1.upto(@seq1_len) do |i|
        1.upto(@seq2_len) do |j|
          if @seq1[i-1]==@seq2[j-1]
            match=1
          else
            match=0
          end
          @lcs_table[i][j] = [ @lcs_table[i-1][j-1] + match, @lcs_table[i-1][j], @lcs_table[i][j-1] ].max
        end
      end
      return @lcs_table[@seq1_len][@seq2_len]
    end
  end
end

q = STDIN.gets.to_i

q.times do
  seq1 = STDIN.gets
  seq2 = STDIN.gets
  seq_pair=SequencePair.new(seq1,seq2) 
  print "#{seq_pair.lcs_len}\n"
end