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 C (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
int main(){
  int YUYUSHIKI,i,j,k;
  char YUKARI[1001],YUZUKO[1001];
  int YUI[1001][1001];
  int MIKAMI,HINATA,NONOMURA;
  scanf("%d",&YUYUSHIKI);
  for(k=0;k<YUYUSHIKI;k++){
    MIKAMI=0;
    scanf("%s%s",YUZUKO,YUKARI);
    HINATA=strlen(YUKARI);
    NONOMURA=strlen(YUZUKO);
    for(i=HINATA;i>0;i--)YUKARI[i]=YUKARI[i-1];
    for(i=NONOMURA;i>0;i--)YUZUKO[i]=YUZUKO[i-1];
    for(i=0;i<=HINATA;i++){
      for(j=0;j<=NONOMURA;j++){
	if(i==0||j==0){
	  YUI[i][j]=0;
	  continue;
	}
	if(YUZUKO[j]==YUKARI[i]){
	  YUI[i][j]=YUI[i-1][j-1]+1;
	  MIKAMI=MIKAMI>YUI[i][j]?MIKAMI:YUI[i][j];
	}else
	  YUI[i][j]=YUI[i][j-1]>YUI[i-1][j]?YUI[i][j-1]:YUI[i-1][j];
      }
    }
    printf("%d\n",MIKAMI);
  }
  
  return 0;
}