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)

# lcs.rb
q = STDIN.gets.split[0].to_i
q.times {
  c=Array.new(1002){Array.new(1002)}
  x = STDIN.gets.split[0]
  y = STDIN.gets.split[0]
  lx = x.length
  ly = y.length


  0.upto(ly) {|i| c[0][i]=0}
  0.upto(lx) {|i| c[i][0]=0}
  max_=0
  for i in 0...lx do 
    for j in 0...ly do
      printf("%d %d %s %s|",i,j,x[i],y[j])
      if x[i]==y[j]
        printf("*")
        c[i+1][j+1] = c[i][j]+1
      elsif c[i][j+1] > c[i+1][j] 
        c[i+1][j+1] = c[i][j+1]
      else
        c[i+1][j+1] = c[i+1][j]
      end
    end
  end

  printf("\n")
  for i in 0..lx do
    for j in 0..ly do
	printf("|%d",c[i][j])
    end
    printf("\n");
  end


  puts c[lx][ly]
}