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(a,b) ( a > b ? a : b)

#define N 1001

int lcs(char *x,char *y);

int main() {
    int n, i;
    char x[N], y[N];

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

int lcs(char *x,char *y) {
    int i, j;
    int lx = strlen(x);
    int ly = strlen(y);
    int LCS[N][N];

    for(i = 0; i <= lx; i++) {
        LCS[i][0] = 0;
    }
    for(j = 1; j <= ly; j++) {
        LCS[0][j] = 0;
    }
    for(i = 1; i <= lx; i++) {
        for(j = 1; j <= ly; j++) {
            if(x[i-1] == y[j-1]) {
                LCS[i][j] = LCS[i-1][j-1] + 1;
            }
            else {
                LCS[i][j] = max(LCS[i-1][j], LCS[i][j-1]);
            }
        }
    }
    return LCS[lx][ly];
}