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)

N = 1000

def max(x, y)
	return x > y ? x : y
end

def lcs(x, y)
	m = x.size
	n = y.size
	maxl = 0
	x = ' ' + x
	y = ' ' + y

	c = []
	for i in 0..N do
		c[i] = []
	end

	for i in 0..m do
		c[i][0] = 0
	end
	for j in 0..n do
		c[0][j] = 0
	end

	for i in 1..m do
		for j in 1..n do
			if x[i] == y[j]
				#debug
				puts "i = #{i}, j = #{j}" if c[i - 1][j - 1] == nil
				c[i][j] = c[i - 1][j - 1] + 1
			else
				c[i][j] = max(c[i - 1][j], c[i][j - 1])
			end
			maxl = max(maxl, c[i][j])
		end
	end

	return maxl
end

n = gets.to_i
n.times do
	s1 = gets.chomp
	s2 = gets.chomp
	puts lcs(s1, s2)
end