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.h>
#include<algorithm>

#define MAX 1000

using namespace std;

int lcs(string s1, string s2) {
    int C[MAX+1][MAX+1];
    int m = s1.size();
    int n = s2.size();
    for (int i = 0; i <= m; ++i) {
        C[i][0] = 0;
    }
    for (int j = 0; j <= n; ++j) {
        C[0][j] = 0;
    }

    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (s1[i-1] == s2[j-1]) {
                C[i][j] = C[i-1][j-1] + 1;
            }
            else {
                C[i][j] = max(C[i][j-1], C[i-1][j]);
            }
        }
    }

    return C[m][n];
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        string s1, s2;
        cin >> s1 >> s2;
        cout << lcs(s1, s2) << endl;
    }
    return 0;
}