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 Rust (judge verdict: Accepted)

use std::io::Read;
use std::cmp::max;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let q: usize = iterator.next().unwrap().parse().unwrap();

    let mut ans = String::new();

    for _ in 0..q {
        let x: Vec<char> = iterator.next().unwrap().chars().collect();
        let y: Vec<char> = iterator.next().unwrap().chars().collect();
        let lcs = lcs(&x, &y);
        ans.push_str(&format!("{}\n", lcs));
    }

    return ans.trim().into();
}

fn lcs(x: &Vec<char>, y: &Vec<char>) -> usize {
    let mut dp: Vec<Vec<usize>> = vec![vec![0; x.len() + 1]; y.len() + 1];

    for i in 1..y.len() + 1 {
        for j in 1..x.len() + 1 {
            dp[i][j] = if x[j - 1] == y[i - 1] {
                dp[i - 1][j - 1] + 1
            } else {
                max(dp[i - 1][j], dp[i][j - 1])
            }
        }
    }

    return dp[y.len()][x.len()];
}