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 JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = Integer.parseInt(sc.nextLine());

		String X, Y;

		for (int i = 0; i < n; i++) {
			X = sc.nextLine();
			Y = sc.nextLine();

			System.out.println(getLcs(X, Y));
		}

		sc.close();
	}

	private static int getLcs(String X, String Y) {
		int m = X.length();
		int n = Y.length();

		char[] x = X.toCharArray();
		char[] y = Y.toCharArray();

		int[][] dp = new int[m + 1][n + 1];

		int max = 0;

		for (int i = 0; i < m + 1; i++) {
			dp[i][0] = 0;
		}

		for (int j = 1; j < n + 1; j++) {
			dp[0][j] = 0;
		}

		for (int i = 1; i < m + 1; i++) {
			for (int j = 1; j < n + 1; j++) {
				if (x[i -1] == y[j -1]) {
					dp[i][j] = dp[i - 1][j - 1] + 1;
				} else {
					dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
				}
				max = Math.max(max, dp[i][j]);
			}
		}

		return max;

	}

}