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>

int hantei(char x[], char y[]) {
int xn = 0, yn = 0, flag = 0, i, j, max = 0;
int c[1001][1001];

while (flag == 0)
{

if (x[xn] != '\0')
{
xn++;
} else {
flag = 1;
}

}

flag = 0;

while (flag == 0) {

if (y[yn] != '\0')
{
yn++;
} else {
flag = 1;
}

}

for (i = 1; i < xn + 1; i++) c[i][0] = 0;
for (j = 1; j < yn + 1; j++) c[0][j] = 0;

for (i = 1; i < xn + 1; i++)
{

for (j = 1; j < yn + 1; 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];
}

if (max < c[i][j]) max = c[i][j];

}

}

return max;
}

int main(void) {
char x[1000], y[1000];
int q, i;

scanf("%d", &q);

for (i = 0; i < q; i++)
{
scanf("%s", x);
scanf("%s", y);
printf("%d\n", hantei(x, y));
}

return 0;
}