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 LongestCommon
  attr_accessor :c

  N = 1000

  def initialize(x, y)
    @x = x
    @y = y
    @c = Array.new(N+1).map { Array.new(N+1, nil) }
    (1..x.size - 1).each { |i| @c[i][0] = 0 }
    (1..y.size - 1).each { |j| @c[0][j] = 0 }
    @c[0][0] = 0
  end

  def lcs
    max = 0
    (1..@x.size - 1).each do |i|
      (1..@y.size - 1).each do |j|
        if @x[i] == @y[j]
          @c[i][j] = @c[i-1][j-1] + 1
        else
          @c[i][j] = [@c[i][j-1], @c[i-1][j]].max
        end
        max = @c[i][j] if @c[i][j] > max
      end
    end
    return max
  end
end

n = gets.to_i
n.times do |i|
  x = gets.strip.chars.to_a.unshift nil
  y = gets.strip.chars.to_a.unshift nil
  lc = LongestCommon.new(x, y)
  puts lc.lcs
end