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 main(void) {
    int num, m, n, i, j;
    int a[MAX], b[MAX];
    char x[MAX], y[MAX], xi;
 
    scanf("%d ", &num);
    while (num--) {
        scanf("%s %s ", x, y);
        m = strlen(x);
        n = strlen(y);
 
        for (i = 0; i < n+1; i++) {
            b[i] = a[i] = 0;
        }
        for (i = 1; i < m+1; i++) {
            xi = x[i-1];
            for (j = 1; j < n+1; j++) {
                if (xi == y[j-1]) {
                    b[j] = a[j-1] + 1;
                } else if (a[j] >= b[j-1]) {
                    b[j] = a[j];
                } else {
                    b[j] = b[j-1];
                }
            }
            for (j = 1; j < n+1; j++) {
                a[j] = b[j];
            }
        }
        printf("%d\n", a[n]);
    }
    return 0;
}