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: Wrong Answer)


def lcs(s1, s2)
  x = s1.length
  y = s2.length
  maxl = 0
  s1 = " " + s1
  s2 = " " + s2
  c = Array.new(1000, Array.new(1000, 0))

  ind_1 = 1
  while ind_1 <= x do
    ind_2 = 1
    while ind_2 <= y
      if x < ind_2
        break
      end
      if s1[ind_1] == s2[ind_2]
        c[ind_1][ind_2] = c[ind_1-1][ind_2-1] + 1
      elsif c[ind_1][ind_2-1] <= c[ind_1-1][ind_2]
        c[ind_1][ind_2] = c[ind_1-1][ind_2]
      else
        c[ind_1][ind_2] = c[ind_1][ind_2-1]
      end
      maxl = [maxl, c[ind_1][ind_2]].max
      ind_2 += 1
    end
    ind_1 += 1
  end
  maxl
end

ans = []
count = gets.chomp.to_i

count.times do
  s1 = gets.chomp
  s2 = gets.chomp
  ans << lcs(s1, s2)
end

puts ans