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 N 1001


        int Comp(int a,int b){
          if(a<b)return b;
          else return a;
        }
        int lcs(char *x,char *y){

          int i,j;
          int lenx,leny;
          int dp[N][N];

          lenx=strlen(x);
          leny=strlen(y);

          for(i=0;i<=lenx;i++)dp[i][0]=0;

          for(j=0;j<=leny;j++)dp[0][j]=0;

          for(i=0;i<lenx;i++){

            for(j=0;j<leny;j++){

              if(x[i]==y[j]) dp[i+1][j+1]=dp[i][j]+1;
              else dp[i+1][j+1]=Comp(dp[i+1][j],dp[i][j+1]);

            }

          }

          return dp[lenx][leny];
        }
        int main(){

          int num;
          char x[N];
          char y[N];

          scanf("%d",&num);
          while(num>0){

            scanf("%s",x);
            scanf("%s",y);

            printf("%d\n",lcs(x,y));
            num--;
          }

          return 0;

        }