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)

def lcs(xs, ys)
  memo = []
  memo[0] = []

  xl = xs.size
  yl = ys.size

  for i in 1..xl
    memo[i] = []
#    memo[i][0] = 0
  end

#  for j in 1..yl
#    memo[0] = []
#    memo[0][j] = 0
#  end

  max = 0
  for i in 1..xl
    for j in 1..yl
      if xs[i-1] == ys[j-1]
        a = memo[i - 1][j - 1]
        memo[i][j] = if a == nil then 1 else a + 1 end
      else
        a = if memo[i - 1][j].nil? then 0 else memo[i - 1][j] end
        b = if memo[i][j - 1].nil? then 0 else memo[i][j - 1] end
        if a >= b
          memo[i][j] = a
        else
          memo[i][j] = b
        end
      end
      if max < memo[i][j]
        max = memo[i][j]
      end
    end
  end
  max
end


n = STDIN.gets.to_i

n.times.each {
  a = STDIN.gets
  b = STDIN.gets
  xs = a.split("")
  ys = b.split("")

  xs.delete_at(xs.size - 1)
  ys.delete_at(ys.size - 1)
  puts lcs(xs, ys)
}