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 <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    cin >> n;
    while(n)
    {
        n--;
        string X,Y;
        int i=0,j=0;
        cin >> X;
        cin >> Y;
        int L[1001][1001];
        for(i=0;i<1001;i++)
            for(j=0;j<1001;j++)
               L[i][j]=0;
        for(i=1;i<=X.size();i++)
            for(j=1;j<=Y.size();j++)
            {
                if(X[i-1]==Y[j-1])
                    L[i][j]=1+L[i-1][j-1];
                else    L[i][j]=max(L[i][j-1],L[i-1][j]);
            }
        cout << L[X.size()][Y.size()] << endl;
    }
    return 0;
}