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 to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let s: Vec<Vec<char>> = (0..2*n).map(|_| read::<String>().chars().collect()).collect();
    //let mut dp: Vec<Vec<u32>> = vec![Vec::new();2*n];
    for x in 0..n {
        //貰うDP
        let mut dp: Vec<Vec<u32>> = vec![vec![0; 1001];1001];
        for i in 0..s[x*2].len() {
            for j in 0..s[x*2+1].len() {
                //println!("s[x*2][{}]:{} s[x*2+1][{}]:{}",i,s[x*2][i],j,s[x*2+1][j]);
                if s[x*2][i] == s[x*2+1][j] {
                    dp[i+1][j+1] = dp[i][j] + 1; 
                }else{
                    dp[i+1][j+1] = std::cmp::max(dp[i+1][j],dp[i][j+1]); 
                }
            }
        }
        println!("{}",dp[s[x*2].len()][s[x*2+1].len()]);
    }
}