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(string1, string2)
	# 扱いやすさの為に1オリジンで扱えるようにする
	# また、配列のほうが処理が早いので変換しておく
	string1 = ("#" + string1).split("")
	string2 = ("#" + string2).split("")

	# 部分的なlcsの結果を格納する表を作る
	# partial_lcs[i][j]はstring1[0..i]とstring2[0..j]のlcsになる
	partial_lcs = Array.new(string1.length) { Array.new(string2.length, 0) }
	1.upto(string1.length - 1) do |i|
		1.upto(string2.length - 1) do |j|

			if string1[i] == string2[j]
				partial_lcs[i][j] = partial_lcs[i - 1][j - 1] + 1

			elsif partial_lcs[i - 1][j] >= partial_lcs[i][j - 1]
				partial_lcs[i][j] = partial_lcs[i - 1][j]

			else
				partial_lcs[i][j] = partial_lcs[i][j - 1]

			end

		end
	end
	partial_lcs[-1][-1] # 表の一番右下、つまり全体のlcs
end

n = STDIN.gets.to_i
n.times do
	string1 = STDIN.gets.chomp!
	string2 = STDIN.gets.chomp!
	puts lcs(string1, string2)
end