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

int main(){
    //input
    long long q;
    cin >> q;
    long long i,j,k, m, n;
    string S, T;
    const char *s = NULL, *t = NULL;
    for(k = 1; k <= q; k++){
        cin >> S;
        m = S.length();
        s = S.c_str();
        cin >> T;
        n = T.length();
        t = T.c_str();
        long long dp[m+1][n+1];
        for(i = 0; i<= m; i++){
            dp[i][0] = 0;
        }
        for(j = 0; j <= n; j++){
            dp[0][j] = 0;
        }
        for(i = 1; i <= m; i++){
            for(j = 1; j <= n; j++){
                dp[i][j] = max({dp[i-1][j-1] + (s[i-1]==t[j-1]) , dp[i-1][j], dp[i][j-1]});
            }
        }
        cout<< dp[m][n] << endl;
    }
    return 0;
}