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)

$t_lcs = []

def calc_lcs(str_x, str_y, i_x, i_y)
  return $t_lcs[i_x][i_y] if $t_lcs[i_x][i_y] >= 0

  if (str_x[i_x] == str_y[i_y]) then
    pre_lcs = calc_lcs(str_x, str_y, i_x-1, i_y-1)
    lcs = pre_lcs + 1
  else
    pre_lcs1 = calc_lcs(str_x, str_y, i_x-1, i_y)
    pre_lcs2 = calc_lcs(str_x, str_y, i_x, i_y-1)
    lcs = [pre_lcs1, pre_lcs2].max
  end

  $t_lcs[i_x][i_y] = lcs
  return $t_lcs[i_x][i_y]
end

def setup_table(str_x, str_y)
  $t_lcs.clear

  (str_x.length).times do 
    cols = []
    (str_y.length).times do
      cols << -1
    end
    $t_lcs << cols
  end

  x_ch = str_x[1]
  y_ch = str_y[1]
  $t_lcs.each_with_index do |row, index|
    next if index == 0
    part_y = str_y[1..index]
    row[1] = (part_y.include?(x_ch)) ? 1 : 0
  end
  $t_lcs[1].each_with_index do |col, index|
    next if index == 0
    part_x = str_x[1..index]
    $t_lcs[1][index] = (part_x.include?(y_ch)) ? 1 : 0
  end
end

def print_table
  $t_lcs.each do |row|
    row.each do |col|
      printf "%3d", col
    end
    puts ""
  end
end


# main
n = gets.chomp.to_i
(n).times do |i|
  str_x = " " + gets.chomp
  str_y = " " + gets.chomp
  setup_table(str_x, str_y)

  if (str_x.length >= str_y.length) then
    diff = str_x.length - str_y.length
    str_y.length.times do |i|
      next if i == 0
      calc_lcs(str_x[0..i+diff], str_y[0..i], i+diff, i)
    end
  else
    diff = str_y.length - str_x.length
    str_x.length.times do |i|
      next if i == 0
      calc_lcs(str_x[0..i], str_y[0..i+diff], i, i+diff)
    end
  end
  puts calc_lcs(str_x, str_y, str_x.length-1, str_y.length-1)
end