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 Scala (judge verdict: Accepted)

import scala.io.StdIn.{readInt,readLine}
import scala.collection.mutable.ArrayBuffer

object Main {
  def max(x:Int,y:Int) = if(x>y) x else y

  def main(args:Array[String]) = {
    val n = readInt
    val b = new ArrayBuffer[Int]

    for(i<-0 until n) {
      val s1 = ' ' + readLine
      val s2 = ' ' + readLine
      val maxN = s1.length
      val maxM = s2.length

      val a = Array.ofDim[Int](maxN,maxM)
      for(i<-1 until maxN) {
        for(j<-1 until maxM) {
          if(s1(i)==s2(j)) {
            a(i)(j) = a(i-1)(j-1)+1
          } else {
            a(i)(j) = max(a(i-1)(j), a(i)(j-1))
          }
        }
      }
      b += a(maxN-1)(maxM-1)
    }
    println(b.mkString("\n"))
  }
}