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.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

class Main {
    public static void main (String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int q = Integer.parseInt(br.readLine());

        StringBuilder sb = new StringBuilder("");

        for (int i = 0; i < q; i++) {
            char[] X = br.readLine().toCharArray();
            char[] Y = br.readLine().toCharArray();
            sb.append(LCS(X, Y));
            sb.append('\n');
        }

        System.out.print(sb);
    }

    private static int LCS(char[] X, char[] Y) {
        int m = X.length;
        int n = Y.length;
        int i, j;
        int[][] len = new int[m+1][n+1];

        for (i = 0; i < m; i++) {
            len[i][0] = 0;
        }
        for (j = 0; j < n; j++) {
            len[0][j] = 0;
        }

        for (i = 1; i <= m; i++) {
            for (j = 1; j <= n; j++) {
                if (X[i-1] == Y[j-1]) {
                    len[i][j] = len[i-1][j-1] + 1;
                } else {
                    len[i][j] = Math.max(len[i][j-1], len[i-1][j]);
                }
            }
        }

        return len[m][n];
    }
}