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: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
void LCS_Length(char* ,char*);
int main()
{
	char A[1001],B[1001];
	int n,i;
	scanf("%d",&n);
	for(i=0;i<n;i++) 
	{
		scanf("%s",A);
		scanf("%s",B);
		LCS_Length(A,B);
	}
	return 0;
}

void LCS_Length(char *X,char *Y)
{
	int m,n;
	int i,j;
	int c[1001][1001];
	m = strlen(X)+1;
	n = strlen(Y)+1;

	for(i=1;i<m;i++) c[i][0] = 0;
	for(j=0;i<n;j++) c[0][j] = 0;

	for(i=1;i<m;i++)
	{
		for(j=1;j<n;j++) 
		{
			if (X[i-1] == Y[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];
			//printf("ans %d %d   %d\n",i,j,c[i][j]);
		}
	}
	printf("%d\n",c[i-1][j-1]);
}