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>
typedef long long ll;
#define rep(i,n) for(ll i=0; i<(ll)(n); i++)
#define max(l,r) ((l)>(r)?l:r)
#define min(l,r) ((l)<(r)?l:r)

int dp[1010][1010] = {0};
int main(){
    int Q;
    scanf("%d", &Q);
    
    rep(q, Q){
        char X[1010], Y[1010];
        scanf("%s%s", X, Y);
        int nx = strlen(X), ny = strlen(Y);
        rep(i, nx+1) dp[i][0] = 0;
        rep(j, ny+1) dp[0][j] = 0;
        rep(i, nx){
            rep(j, ny){
                dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
                if(X[i] == Y[j]) dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + 1);  
            }
        }
        printf("%d\n", dp[nx][ny]);
    }
}