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++11 (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define rep(i,x) for(int i=0;i<x;++i)
#define rrep(i,x) for(int i=x-1;i>=0;--i)

int dp[1010][1010];

int main()
{
    int q; cin >> q;

    rep(_, q) {
        memset(dp, -1, sizeof(dp));
        dp[0][0] = 0;

        string a,b; cin >> a>>b;

        int n = a.size(), m = b.size();

        a += '$';
        b += '!';

        rep(i, n+1) {
            rep(j, m+1) {
                dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
                dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);

                if (a[i] == b[j]) {
                    dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + 1);
                }
            }
        }

        cout << dp[n][m] << endl;
    }
}