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::cmp;
use std::io;
use std::io::prelude::*;

fn solve(dp: &mut Vec<Vec<usize>>, x: &str, y: &str) -> usize {
    for (i, xi) in x.chars().enumerate() {
        for (j, yj) in y.chars().enumerate() {
            dp[i + 1][j + 1] = if xi == yj {
                dp[i][j] + 1
            } else {
                cmp::max(dp[i + 1][j], dp[i][j + 1])
            };
        }
    }
    dp[x.len()][y.len()]
}

fn main() {
    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let n = lines.next().unwrap().unwrap().parse().unwrap();

    for _ in 0..n {
        let x = lines.next().unwrap().unwrap();
        let y = lines.next().unwrap().unwrap();
        let nx = x.len();
        let ny = y.len();
        let mut dp = vec![vec![0; ny + 1]; nx + 1];

        println!("{}", solve(&mut dp, &x, &y));
    }
}