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>
#include <stdlib.h>
#include <math.h>
#define SIZE 1010


char* reverseString(char* str);
int max(int a, int b);

int main(void) {
    char str1[SIZE], str2[SIZE];
    int str1len, str2len, i, j, k, anscount = 0;
    int dp[SIZE][SIZE];
    char ansstr[SIZE];
    int q;
    scanf("%d\n",&q);
    for (k = 0; k < q; k++) {
        memset(dp,0,sizeof(dp));

        fgets(str1,SIZE,stdin);
        fgets(str2,SIZE,stdin);
        str1len = strlen(str1);
        str2len = strlen(str2);

        if (str1[str1len - 1] == '\n') {
            str1[--str1len] = '\0';
        }
        if (str2[str2len - 1] == '\n') {
            str2[--str2len] = '\0';
        }

        for (i = 0; i < str1len; i++) {
            for (j = 0; j < str2len; j++) {
                if (str1[i] == str2[j]) {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                } else {
                    dp[i + 1][j + 1] = max(dp[i][j + 1],dp[i + 1][j]);
                }
            }
        }

        printf("%d\n",dp[str1len][str2len]);
    }

    return 0;
}

char* reverseString(char* str) {
    char* ansstr;
    int len, i;

    len = strlen(str);
    ansstr = (char*)malloc(sizeof(char) * len);
    
    for (i = 0; i < len; i++) {
        ansstr[i] = str[len - i - 1];
    }
    return ansstr;
}

int max(int a, int b) {
    return a > b ? a : b;
}