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>
#include <stdlib.h>
#define MAX 1001


int main(void){
	int i,j;
	char s1[MAX],s2[MAX];
	int n;
	
	scanf("%d",&n);
	
	for(i=0; i<n; i++){
		scanf("%s %s",s1,s2);
		int l1 = strlen(s1);
		int l2 =strlen(s2);
		for(j=l1; j >= 0; j--) s1[j+1]=s1[j];
		s1[0] =' ';
		for(j=l2; j >= 0; j--) s2[j+1]=s2[j];
		s2[0] =' ';
//		printf("%s %s\n",s1,s2);
		printf("%d\n", lcs(s1,s2) );
		}
	return 0;
}

int lcs(char s1[],char s2[]){
	int i,j;
	int m = strlen(s1);
	int n = strlen(s2);
	
	/*
	int **c;
	 c = (int **)malloc(sizeof(int *)*(m+1));
	for(i=0;i<=n;i++){
		c[i] = (int *)malloc(sizeof(int)*(n+1));
	}
	*/
	int c[MAX+1][MAX+1];
//	printf("debug1");
	for(i=0; i<=m; i++) c[i][0] = 0;
	for(i=1; i<=n; i++) c[0][i] = 0;
	
	
	
	for(i=1; i<=m; i++){
//			printf("debug2\n");
		for(j=1; j<=n; j++){
//				printf("debug3\n");
			if(s1[i] == s2[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];
			}
		}
	}
//		printf("debug4\n");


	return c[i-2][j-2];
}