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++ (judge verdict: Accepted)

#include <iostream>
using namespace std;

const int MAX_LEN = 1000;

int lcs(string x, string y) {
  int a[MAX_LEN + 1][MAX_LEN + 1];
  for (int i = 0; i <= MAX_LEN; i++)
    a[i][0] = a[0][i] = 0;

  for (int i = 1; i <= x.size(); i++) {
    for (int j = 1; j <= y.size(); j++) {
      if (x[i-1] == y[j-1]) {
        a[i][j] = a[i-1][j-1] + 1;
      } else if (a[i][j-1] > a[i-1][j]) {
        a[i][j] = a[i][j-1];
      } else {
        a[i][j] = a[i-1][j];
      }
    }
  }
  return a[x.size()][y.size()];
}

int main() {
  string x, y;
  int q; cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> x >> y;
    cout << lcs(x, y) << endl;
  }
  return 0;
}