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::BufRead;

pub struct StdinReader<R: BufRead> {
    pub buf_reader: R,
    pub buf: String,
}
impl<R: BufRead> StdinReader<R> {
    pub fn new(r: R) -> Self {
        StdinReader {
            buf_reader: r,
            buf: String::with_capacity(10000),
        }
    }
}

#[macro_export]
macro_rules! input {
    ($reader : expr, $t : ty) => { // one variable
        match &mut $reader {
            reader => {
                use std::io::BufRead;
                reader.buf_reader.read_line(&mut reader.buf).unwrap();
                let tmp_var: $t = reader.buf.trim().parse().unwrap(); 
                reader.buf.clear();
                tmp_var
            },
        }
    };
    ($reader : expr, $t0 : ty, $t1 : ty) => { // 2 tuple
        match &mut $reader {
            reader => {
                use std::io::BufRead;
                reader.buf_reader.read_line(&mut reader.buf).unwrap();
                let tmp_var: Vec<$t0> = reader.buf.trim().split_whitespace().map(|a| a.parse().unwrap()).collect();
                reader.buf.clear();
                (tmp_var[0], tmp_var[1])
            }
        }
    };
    ($reader : expr, $t0 : ty, $t1 : ty, $t2 : ty) => { // 3 tuple
        match &mut $reader {
            reader => {
                use std::io::BufRead;
                reader.buf_reader.read_line(&mut reader.buf).unwrap();
                let tmp_var: Vec<$t0> = reader.buf.trim().split_whitespace().map(|a| a.parse().unwrap()).collect();
                reader.buf.clear();
                (tmp_var[0], tmp_var[1], tmp_var[2])
            }
        }
    };
    ($reader : expr, $t : ty; $size : expr) => { // Vector
        match &mut $reader {
            reader => {
                use std::io::BufRead;
                reader.buf_reader.read_line(&mut reader.buf).unwrap();
                let tmp_var: $t = reader.buf.trim().split_whitespace().map(|a| a.parse().unwrap()).collect();
                reader.buf.clear();
                tmp_var
            }
        }
    };
}

fn main() {
    let handle = std::io::stdin();
    let mut r = StdinReader::new(handle.lock());

    let q = input!(r, usize);
    for _ in 0..q {
        let s: Vec<char> = input!(r, String).chars().collect();
        let t: Vec<char> = input!(r, String).chars().collect();
        let mut dp: Vec<Vec<usize>> = vec![vec![0; t.len()+1]; s.len()+1];

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