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 OCaml (judge verdict: Accepted)

let lcs (x, y) =
  let n = String.length x in
  let m = String.length y in
  let dp = Array.make_matrix (n + 1) (m + 1) 0 in
  for i = 1 to n do
    for j = 1 to m do
      dp.(i).(j) <-
        if x.[i-1] = y.[j-1] then dp.(i-1).(j-1) + 1
        else
          let a = dp.(i-1).(j) in
          let b = dp.(i).(j-1) in
          if a > b then a else b;
    done
  done;
  dp.(n).(m)

let () =
  let q = read_int () in
  for _ = 0 to q - 1 do
    Scanf.scanf "%s\n%s\n" (fun x y -> (x, y))
    |> lcs
    |> Printf.printf "%d\n"
  done