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

public class Main {

  public static void main(String[] args) {
    
    new Main().run();
  }

  public void run() {

    try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

      int q = Integer.parseInt(br.readLine());
      StringBuilder buf  = new StringBuilder();

      for(int i=0; i<q; i++) {
        buf.append( deriveLcs(br.readLine(), br.readLine()) )
           .append("\n");
      }

      System.out.print(buf);
    }
    catch (IOException e) {
      System.out.println("IOException!");
    }
  }

  private int deriveLcs(String seqX, String seqY) {
    int m = seqX.length();
    int n = seqY.length();
    int size = m > n ? m+1 : n+1;
    int[][] subSeq = new int[size][size];
    int maxSubSeq = 0;

    for(int i=1; i<=m; i++) subSeq[i][0] = 0;
    for(int j=1; j<=n; j++) subSeq[0][j] = 0;

    for(int i=1; i<=m; i++) {
      for(int j=1; j<=n; j++) {
        if(seqX.charAt(i-1) == seqY.charAt(j-1))
          subSeq[i][j] = subSeq[i-1][j-1] + 1;
        else
          subSeq[i][j] = Math.max(subSeq[i-1][j], subSeq[i][j-1]);

        maxSubSeq = Math.max(maxSubSeq, subSeq[i][j]);
      }
    }

    return maxSubSeq;
  }
}