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)

def lcs(str1, str2)
  arr = Array.new(str1.size + 1) { Array.new(str2.size + 1, 0) }

  max = 0

  m = str1.size
  n = str2.size

  str1 = ' ' + str1
  str2 = ' ' + str2

  1.upto(m) do |i|
    1.upto(n) do |j|
      if str1[i] == str2[j]
        arr[i][j] = arr[i - 1][j - 1] + 1
      else
        arr[i][j] = [arr[i - 1][j], arr[i][j - 1]].max
      end

      max = [arr[i][j], max].max
    end
  end

  max
end

if $PROGRAM_NAME == __FILE__
  n = gets.to_i

  n.times do
    str1 = gets.chomp
    str2 = gets.chomp

    puts lcs(str1, str2)
  end
end