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)

fn main() {
    let n = read();
    for _ in 0..n {
        let s: Vec<char> = read::<String>().chars().collect();
        let t: Vec<char> = read::<String>().chars().collect();
        // [sの文字目][tの文字目]
        let mut memo = vec![vec![(0, (0, 0)); t.len() + 1]; s.len() + 1];

        for i in 0..s.len() {
            for j in 0..t.len() {
                if s[i] == t[j] {
                    if memo[i][j].0 + 1 > memo[i + 1][j + 1].0 {
                        memo[i + 1][j + 1] = (memo[i][j].0 + 1, (i, j));
                    }
                } else {
                    if memo[i][j + 1].0 < memo[i + 1][j].0 {
                        memo[i + 1][j + 1] = (memo[i + 1][j].0, (i + 1, j))
                    } else {
                        memo[i + 1][j + 1] = (memo[i][j + 1].0, (i, j + 1))
                    }
                }
            }
        }
        println!("{}", memo[s.len()][t.len()].0);
    }
}

// =========
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet};
#[allow(unused_imports)]
use std::process::exit;

#[allow(dead_code)]
const MOD: usize = 1000000007;

fn read<T: std::str::FromStr>() -> T {
    use std::io::Read;
    let stdin = std::io::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")
}

// =========