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 JavaScript (judge verdict: Time Limit Exceeded)

//-----

var log = console.log;
if (typeof process != "undefined") {
	var input = "";
	process.stdin.resume();
	process.stdin.setEncoding('utf8');
	process.stdin.on('data', function (chunk) {
		input += chunk;
	});
	process.stdin.on('end', function () {
		var lines = input.split("\n");
		main(lines);
	});
}

//-----

function main(lines) {
	lines.shift();
	while (lines.length > 0) {
	try{
		var dataset = lines.slice(0, 2);
		dataset = dataset.map(function (row) {
			return row.split("").filter(function (e) {
				var code = e.charCodeAt(0);
				return 97 <= code && code <= 122;
			}).join("");
		});
		processDataset(dataset);
		lines = lines.slice(2);
	}catch(e){
		log(e);
	}
	}
}

function processDataset(dataset) {
	var A = [];
	var f = dataset[0];
	var s = dataset[1];
	for (var i = 0; i < f.length; i++) {
		A[i] = [];
		for (var j = 0; j < s.length; j++) {
			if (f.charAt(i) == s.charAt(j)) {
				A[i][j] = 1;
			} else {
				A[i][j] = 0;
			}
			if (i == 0 && j == 0) {
				A[i][j] += 0;
			} else if (i == 0) {
				A[i][j] = Math.max(A[i][j - 1], A[i][j]);
			} else if (j == 0) {
				A[i][j] = Math.max(A[i - 1][j], A[i][j]);
			} else {
				A[i][j] = Math.min(Math.max(A[i - 1][j], A[i][j - 1]) + A[i][j], Math.min(i, j) + 1);
			}
		}
	}
	log(A[i - 1][j - 1]);
}