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 C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

int solve(string &s, string &t, vector<vector<int>> &memo, int i = 0,
          int j = 0) {
  if (i >= s.size() || j >= t.size()) {
    return 0;
  }

  if (memo[i][j] < 0) {
    if (*(s.end() - i - 1) == *(t.end() - j - 1)) {
      memo[i][j] = solve(s, t, memo, i + 1, j + 1) + 1;
    } else {
      memo[i][j] =
          max(solve(s, t, memo, i + 1, j), solve(s, t, memo, i, j + 1));
    }
  }

  return memo[i][j];
}
int main(int argc, char *argv[]) {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    string s, t;
    cin >> s >> t;
    vector<vector<int>> memo(s.size(), vector<int>(t.size(), -1));
    cout << solve(s, t, memo) << '\n';
  }

  return 0;
}