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::{self, Read};
use std::str::FromStr;

pub struct Scanner<R: Read> {
    reader: R,
}
impl<R: Read> Scanner<R> {
    pub fn new(reader: R) -> Scanner<R> {
        Scanner { reader: reader }
    }
    pub fn read<T: FromStr>(&mut self) -> T {
        let s = self
            .reader
            .by_ref()
            .bytes()
            .map(|c| c.expect("failed to read char") as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        s.parse::<T>().ok().expect("failed to parse token")
    }
}

fn lcs(x: &String, y: &String) -> u64 {
    // なお、x.len>y.lenであるものとする
    let xlen = x.len();
    let ylen = y.len();
    assert!(xlen >= ylen);

    // 方針: 解説より
    // xi == yi のとき、そのLCSはx[..i]とy[..j]のLCSにxi(=yi)を加えたもの
    // xi != yi のとき、そのLCSはx[..i+1]とy[..j] or x[..i]とy[..j+1]のLCSの長い方となる

    let xc: Vec<char> = x.chars().collect();
    let yc: Vec<char> = y.chars().collect();

    // dp[i][j] をx[0..i+1], y[0..j+1]の最長共通部分列の長さとする
    // X = {x1, x2, ..., xn}, Y = {y1,y2,...,ym}としたとき、
    // dp[i][j]がX[..i+1], Y[..j+1]のLCSの長さとする (つまり添え字0は0埋め)
    let mut dp: Vec<Vec<u64>> = vec![vec![0; ylen+1]; xlen+1];

    for i in 1..xlen+1 {
        for j in 1..ylen+1 {
            if xc[i-1] == yc[j-1] {
                dp[i][j] = dp[i-1][j-1] + 1;
            } else {
                dp[i][j] = if dp[i-1][j] > dp[i][j-1] {
                    dp[i-1][j]
                } else {
                    dp[i][j-1]
                };
            }
        }
    } 

    dp[xlen][ylen]
}

fn main() {
    let sin = io::stdin();
    let sin = sin.lock();
    let mut sc = Scanner::new(sin);

    let n: usize = sc.read();
    for _ in 0..n {
        let x: String = sc.read();
        let y: String = sc.read();
        let res = if x.len() > y.len () {
            lcs(&x, &y) 
        } else {
            lcs(&y, &x)
        };
        println!("{}", res);
    }
}