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 lcs(char *,char *);

int main(){
	char S1[20],S2[20];
	int i,n;
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%s",S1);
		scanf("%s",S2);
		int c[1000][1000];
    int max=0;
    int l1,l2,a,b;
    l1=strlen(S1)+1;
    l2=strlen(S2)+1;
    for(a=0;a<l1;a++)c[0][a]=0;
    for(a=0;a<l2;a++)c[a][0]=0;

    for(a=1;a<l1;a++){
    	for(b=1;b<l2;b++){
    		if(S1[a-1]==S2[b-1])c[a][b]=c[a-1][b-1]+1;
    		else if(c[a-1][b]>=c[a][b-1])c[a][b]=c[a-1][b];
    		else c[a][b]=c[a][b-1];
    	}
    }

		//max=lcs(S1,S2);
		printf("%d\n",c[l1-1][l2-1]);
	}
	return 0;
}

int lcs(char *x,char *y){
	int c[1000][1000];
	int i,j,m,n;
	m=strlen(x);
	n=strlen(y);
	int max=0;
  x=strcat(" ",x);
  y=strcat(" ",y);
  for(i=1;i<=m;i++)c[i][0]=0;
	for(j=1;j<=n;j++)c[0][j]=0;

	for(i=1;i<=m;i++){
		for(j=1;j<=n;j++){
			if(x[i]==y[j]){
				c[i][j]=c[i-1][j-1]+1;
			}
			else{
				if(c[i-1][j]>c[i][j-1]) c[i][j]=c[i-1][j];
				else c[i][j]=c[i][j-1];
			}
			if(max<c[i][j]) max=c[i][j];
		}
	}
	return max;
}*/
#include<stdio.h>
#include<string.h>
int main()
{
  int q,i,j;
  char S1[1001],S2[1001];
  scanf("%d",&q);
  for(i=0;i<q;i++){
    scanf("%s",S1);
    scanf("%s",S2);
    printf("%d\n",LCS(S1,S2));
  }
  return 0;
}

int LCS(char *S1,char *S2)
{
  int c[1001][1001],i,j,x,y;
  x=strlen(S1)+1;
  y=strlen(S2)+1;
  for(i=1;i<x;i++)c[0][i]=0;
  for(i=0;i<y;i++)c[i][0]=0;

  for(i=1;i<x;i++){
    for(j=1;j<y;j++){
      if(S1[i-1]==S2[j-1])c[i][j]=c[i-1][j-1]+1;
      else if(c[i-1][j]>=c[i][j-1])c[i][j]=c[i-1][j];
      else c[i][j]=c[i][j-1];
    }
  }
  return c[x-1][y-1];
}