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 JavaScript (judge verdict: Accepted)

(function main() {
  const lines = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
  const n = Number(lines.shift());
  const MAX = 1000;
  const c = Array.from(Array(MAX + 1), () => []);

  let i, j, X, Y, xlen, ylen, maxlen;
  while ((X = lines.shift()) && (Y = lines.shift())) {
    xlen = X.length;
    ylen = Y.length;
    for (i = 0; i <= xlen; i++) c[i][0] = 0;
    for (j = 0; j <= ylen; j++) c[0][j] = 0;

    for (i = 1; i <= xlen; i++) {
      for (j = 1; j <= ylen; j++) {
        if (X[i - 1] === Y[j - 1]) c[i][j] = c[i - 1][j - 1] + 1;
        else c[i][j] = Math.max(c[i - 1][j], c[i][j - 1]);
      }
    }
    console.log(c[xlen][ylen]);
  }
})();