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(input) {
    var l = 1;
    input = input.trim().split("\n").map(function(x) { return x.split(" ")});    
    var q = parseInt(input[0][0], 10);

    for (var k = 0; k < q; k++){
        var X = input[l][0].split("");
        var Y = input[l + 1][0].split("");

        var arr_dp = [];
        for (var j = 0; j <= Y.length; j++){
            arr_dp.push([]);
            arr_dp[j][0] = 0;
        }
        for (var i = 0; i <= X.length; i++){
            arr_dp[0][i] = 0;
        }
        for (var j = 0; j < Y.length; j++){
            for (var i = 0; i < X.length; i++){
                if (X[i] === Y[j]) arr_dp[j + 1][i + 1] = arr_dp[j][i] + 1;
                else arr_dp[j + 1][i + 1] = Math.max(arr_dp[j + 1][i], arr_dp[j][i + 1]);
            }
        }
        console.log(arr_dp[j][i]);
        l+= 2;
    }
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));