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 PHP (judge verdict: Time Limit Exceeded)

<?php
// Dynamic Planning
solve2();

// 模範解
function solve2(){
	$q = (int)trim(fgets(STDIN));
	for ($z=0; $z < $q; $z++) { 
		$s1 = trim(fgets(STDIN));
		$l1 = strlen($s1);
		$s2 = trim(fgets(STDIN));
		$l2 = strlen($s2);
		$dp = [];
		for ($i=0; $i <= $l1; $i++) { 
			$dp[$i][0] = 0;
		}
		for ($i=1; $i <= $l2; $i++) { 
			$dp[0][$i] = 0;
		}
		for ($i=1; $i <= $l1; $i++) { 
			for ($j=1; $j <= $l2; $j++) { 
				if ($s1[$i-1] === $s2[$j-1]) {
					$dp[$i][$j] = $dp[$i-1][$j-1] + 1;
				}elseif ($dp[$i-1][$j] >= $dp[$i][$j-1]) {
					$dp[$i][$j] = $dp[$i-1][$j];
				}else{
					$dp[$i][$j] = $dp[$i][$j-1];
				}
			}
		}
		$max = 0;
		for ($i=0; $i <= $l1; $i++) { 
			for ($j=0; $j <= $l2; $j++) { 
				$max = max($max, $dp[$i][$j]);
			}
		}
		echo $max . PHP_EOL;
	}
}