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: Accepted)

def lcs(x, y)
  m = x.length
  n = y.length
  # maxl = 0
  c = Array.new(m+1) { Array.new(n+1, 0) }
  for i in 1..m
    for j in 1..n
      if x[i-1] == y[j-1]
        c[i][j] = c[i - 1][j - 1] + 1
      elsif c[i - 1][j] >= c[i][j - 1]
        c[i][j] = c[i - 1][j]
      else
        c[i][j] = c[i][j - 1]
      end
      # maxl = [maxl, c[i][j]].max
      # puts c.join(' ')
      # puts "c[1][3] : #{c[1][3]}"
      # puts "x: #{x[i]}, `y: #{y[j]}, i: #{i}, j: #{j}, c[i][j]: #{c[i][j]}, c[i-1][j-1]: #{c[i-1][j-1]}, maxl: #{maxl}"
    end
  end
  c[m][n]
end

gets.to_i.times do
  x = gets.chomp.chars
  y = gets.chomp.chars
  puts lcs(x, y)
end