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)

#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
#[allow(unused_imports)]
use std::cmp::{max, min, Ordering};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};
#[allow(unused_imports)]
use std::iter::FromIterator;
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[macro_export]
macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut parser = Parser :: from_str ( $ s ) ; input_inner ! { parser , $ ( $ r ) * } } ; ( parser = $ parser : ident , $ ( $ r : tt ) * ) => { input_inner ! { $ parser , $ ( $ r ) * } } ; ( new_stdin_parser = $ parser : ident , $ ( $ r : tt ) * ) => { let stdin = std :: io :: stdin ( ) ; let reader = std :: io :: BufReader :: new ( stdin . lock ( ) ) ; let mut $ parser = Parser :: new ( reader ) ; input_inner ! { $ parser , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { input ! { new_stdin_parser = parser , $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! input_inner { ( $ parser : ident ) => { } ; ( $ parser : ident , ) => { } ; ( $ parser : ident , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ parser , $ t ) ; input_inner ! { $ parser $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! read_value { ( $ parser : ident , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ parser , $ t ) ) ,* ) } ; ( $ parser : ident , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ parser , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ parser : ident , chars ) => { read_value ! ( $ parser , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ parser : ident , usize1 ) => { read_value ! ( $ parser , usize ) - 1 } ; ( $ parser : ident , $ t : ty ) => { $ parser . next ::<$ t > ( ) . expect ( "Parse error" ) } ; }
use std::io;
use std::io::BufRead;
use std::str;
pub struct Parser<R> {
    reader: R,
    buf: Vec<u8>,
    pos: usize,
}
impl Parser<io::Empty> {
    pub fn from_str(s: &str) -> Parser<io::Empty> {
        Parser {
            reader: io::empty(),
            buf: s.as_bytes().to_vec(),
            pos: 0,
        }
    }
}
impl<R: BufRead> Parser<R> {
    pub fn new(reader: R) -> Parser<R> {
        Parser {
            reader: reader,
            buf: vec![],
            pos: 0,
        }
    }
    pub fn update_buf(&mut self) {
        self.buf.clear();
        self.pos = 0;
        loop {
            let (len, complete) = {
                let buf2 = self.reader.fill_buf().unwrap();
                self.buf.extend_from_slice(buf2);
                let len = buf2.len();
                if len == 0 {
                    break;
                }
                (len, buf2[len - 1] <= 0x20)
            };
            self.reader.consume(len);
            if complete {
                break;
            }
        }
    }
    pub fn next<T: str::FromStr>(&mut self) -> Result<T, T::Err> {
        loop {
            let mut begin = self.pos;
            while begin < self.buf.len() && (self.buf[begin] <= 0x20) {
                begin += 1;
            }
            let mut end = begin;
            while end < self.buf.len() && (self.buf[end] > 0x20) {
                end += 1;
            }
            if begin != self.buf.len() {
                self.pos = end;
                return str::from_utf8(&self.buf[begin..end]).unwrap().parse::<T>();
            } else {
                self.update_buf();
            }
        }
    }
}
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
const BIG_STACK_SIZE: bool = true;
#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}
fn solve() {
    input! {
        Q: usize,
        X: [(chars,chars); Q],
    }

    for x in &X {
        // dbg!(x);
        let ref s1 = x.0;
        let ref s2 = x.1;
        let mut dp = vec![vec![0; s2.len()+1]; s1.len()+1];
        for i in 1..s1.len()+1 {
            for j in 1..s2.len()+1 {
                dp[i][j] = if s1[i-1] == s2[j-1] {
                    dp[i-1][j-1] + 1
                } else {
                    max(dp[i][j-1], dp[i-1][j])
                };
            }
        }
        println!("{}", dp[s1.len()][s2.len()]);
    }
}