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 1000
char s1[MAX],s2[MAX];

int Size(char *x){
  int i=0;
  while(x[i]!='\0'){
    i++;
  }
  // printf("%d\n",i);
  return i;
}

void push1(int m){
  char p[MAX+1];
  int i;
  for(i=0;i<m;i++)p[i]=s1[i];
   s1[0]=' ';
  for(i=1;i<m+1;i++)s1[i]=p[i-1];
}

  void push2(int m){
  char p[MAX+1];
  int i;
  for(i=0;i<m;i++)p[i]=s2[i];
   s2[0]=' ';
  for(i=1;i<m+1;i++)s2[i]=p[i-1];
}

int max(int x,int y){
  if(x>y)return x;
  else return y;
}
int lcs(void){
  int c[MAX+1][MAX+1];
  int m,n,i,j,maxl=0;
  // m=sizeof(s1)/sizeof(s1[0]);
  // n=sizeof(s2)/sizeof(s2[0]);

  m=Size(s1);
  n=Size(s2);
  
  push1(m);
  push2(n);

  for(i=1;i<=m;i++){
    for(j=1;j<=n;j++){
      if(s1[i]==s2[j]){
	c[i][j]=c[i-1][j-1]+1;
      }else {
	c[i][j]=max(c[i-1][j],c[i][j-1]);
      }
      maxl=max(maxl,c[i][j]);
    }
  }
  return maxl;
}


int main(){
 
  int n,i;

  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%s",s1);
    scanf("%s",s2);
    printf("%d\n",lcs());
  }

  return 0;
  
}