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 N 1000

void cm(char a[] , char b[]);
int max(int a , int b);
int c[N+1][N+1];

int main()
{
	char a[N];
	char b[N];
	int i , n;
	scanf("%d",&n);
	for(i = 1; i <= n; i++)
	{
		scanf("%s%s",a+1,b+1);
		a[0]=b[0]= ' ';
		cm(a,b);
	}
}

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

void cm(char  a[]  , char b[])
{
	int i , j;
	int m1,m2;
	m1 = strlen(a)-1;
	m2 = strlen(b)-1;
	int ma = 0;
	for(i=0;i<m1;i++)
	{
		c[0][i]=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(a[i]==b[j])
			{
				c[i][j]=c[i-1][j-1]+1;
			}
			else
			{
				c[i][j]=max(c[i-1][j],c[i][j-1]);
			}
			ma = max(ma,c[i][j]);
				
		}
	}
	printf("%d\n",ma);
}