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::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed reading char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed parsing")
}

fn main() {
    let q = read();
    let mut xs: Vec<String> = Vec::new();
    let mut ys: Vec<String> = Vec::new();
    for _ in 0..q {
        xs.push(read());
        ys.push(read());
    }

    for n in 0..q {
        let s: Vec<char> = xs[n].chars().collect();
        let t: Vec<char> = ys[n].chars().collect();
        // dp[i][j] = sの長さiの部分文字列と、tの長さjの部分文字列に対するLCS
        let mut dp = vec![vec![0; t.len() + 1]; s.len() + 1];

        for i in 1..s.len() + 1 {
            for j in 1..t.len() + 1 {
                dp[i][j] =
                    if s[i - 1] == t[j - 1] {
                        dp[i - 1][j - 1] + 1
                    } else {
                        std::cmp::max(dp[i - 1][j], dp[i][j - 1])
                    };
            }
        }
        println!("{}", dp[s.len()][t.len()]);
    }
}