Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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++14 (judge verdict: Accepted)

// C++ 14
#include <bits/stdc++.h>
#include <iostream>
#include <string>
#include <vector>
#include <array>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>

#define ll long long
#define loop(x, start, end) for(int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)

#define MAX 1000

using namespace std;



int lcs(string &s1, string &s2) {
    int dp[MAX+1][MAX+1];
    int l1 = s1.length();
    int l2 = s2.length();
    int maxl = 0;
    
    s1 = ' ' + s1;
    s2 = ' ' + s2;
    
    loop(i, 0, l1+1) {
        dp[i][0] = 0;
    }
    loop(j, 0, l2+1) {
        dp[0][j] = 0;
    }
    
    loop (i, 1, l1+1) {
        loop (j, 1, l2+1) {
            if (s1[i] == s2[j]) dp[i][j] = 1 + dp[i-1][j-1];
            else dp[i][j] = max(
                dp[i-1][j],
                dp[i][j-1]
            );
            
            maxl = max(maxl, dp[i][j]);
        }
        
    }
    
    return maxl;
}

int main(void) {
    int n;
    string s1, s2;
    
    cin >> n;
    loop(i, 0, n) {
        cin >> s1 >> s2;
        cout << lcs(s1, s2) << endl;
    }
}