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 Kotlin (judge verdict: Accepted)

fun main(args: Array<String>) {
    val MAX = 1000
    var N = Integer.parseInt(readLine());
    var L = Array(MAX+1) { _ -> IntArray(MAX+1) };
    
    for (x in 1..N) {
        val a = readLine();
        val b = readLine();
        
        val n = a!!.length;
        val m = b!!.length;
        
        for (i in 0..(n-1)) {
            for (j in 0..(m-1)) {
                if (a[i] == b[j]) {
                    L[i+1][j+1] = L[i][j] + 1
                } else {
                    L[i+1][j+1] = if (L[i+1][j] > L[i][j+1]) {
                        L[i+1][j]
                    } else {
                        L[i][j+1]
                    }
                }
            }
        }
        println(L[n][m]);
    }
}