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>
#define MAX 1000
void compare(char s1[],char s2[]);
int max(int a,int b);
int c[MAX+1][MAX+1];

int main(void){
	char s1[MAX];
	char s2[MAX];
	int i,n;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
	scanf("%s%s",s1+1,s2+1);
		s1[0]=s2[0]=' '; //空白を代入する
		compare(s1,s2);
	}
}

int max(int a,int b){
	if(a>b){
		return a;
	}
	return b;
}

void compare(char s1[],char s2[]){
	int i,j;
	int m1,m2;
	m1=strlen(s1)-1;
	m2=strlen(s2)-1;
	
	int maximum=0;
	
	for(i=0;i<m1;i++){
		c[i][0]=0;
	}
	for(j=0;j<m2;j++){
		c[0][j]=0;
	}
	for(int i=1;i<=m1;i++){
		for(int j=1;j<=m2;j++){
			if(s1[i]==s2[j]){
				c[i][j]=c[i-1][j-1]+1;
			}
			else {
				c[i][j]=max(c[i-1][j],c[i][j-1]);
			}
			maximum=max(maximum,c[i][j]);
		}
	}
	
	printf("%d\n",maximum);
}