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

  int i,j;
  int xlen,ylen;
  int dp[1001][1001];
  
  xlen=strlen(x);
  ylen=strlen(y);
  for(i=0;i<=xlen;i++)dp[i][0]=0;
  for(j=0;j<=ylen;j++)dp[0][j]=0;
  for(i=0;i<xlen;i++){
    for(j=0;j<ylen;j++){
      if(x[i]==y[j]) dp[i+1][j+1]=dp[i][j]+1;
      else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
    }
  }
  return dp[xlen][ylen];
}
int main(){
  int q;
  char x[1001];
  char y[1001];
  int leng;
  scanf("%d",&q);
  while(q>0){
    scanf("%s",x);
    scanf("%s",y);

    printf("%d\n",lcs(x,y));
    q--;
  }
  return 0;
}