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 <stdio.h>
#include <string.h>
#define MAX 1001
 
int LCS();

char X[MAX], Y[MAX];
 

int main() {
    int n, i;
 
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%s%s", X, Y);
        printf("%d\n", LCS());
    }
 
    return 0;
}

int LCS() {
    int i, j;
    int m = strlen(X);
    int n = strlen(Y);
    int c[m + 1][n + 1];
 
    for (i = 1; i <= m; i++) c[i][0] = 0;
    for (j = 0; j <= n; j++) c[0][j] = 0;
 
    for (i = 1; i <= m; i++) {
        for (j = 1; j <= n; j++) {
            if (X[i - 1] == Y[j - 1]) {
                c[i][j] = c[i - 1][j - 1] + 1;
            } else if (c[i - 1][j] >= c[i][j - 1]) {
                c[i][j] = c[i - 1][j];
            } else {
                c[i][j] = c[i][j - 1];
            }
        }
    }
 
    return c[m][n];
}