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<math.h>

#define MAX 150
#define LEN 1000

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

int lcs(char *x, char *y){
  int i, j, lenx, leny;
  int c[LEN + 1][LEN + 1] = {};
  lenx = strlen(x);
  leny = strlen(y);

  //  0 A C D B A
  //0 0 0 0 0 0 0
  //A 0 1 1 1 1 1
  //C 0 1 2 2 2 2
  //B 0 1 2 2 3 3<-
  
  for(i = 0; i < lenx; i++){
    for(j = 0; j < leny; j++){
       if(x[i] == y[j])
	c[i + 1][j + 1] = c[i][j] + 1;
      else
	c[i + 1][j + 1] = max(c[i + 1][j], c[i][j + 1]);
    }
  }

  return c[lenx][leny];
}

int main(){
  int i,q;
  char x[LEN],y[LEN];

  
  scanf("%d",&q);
  for(i = 0; i < q; i++){
    scanf("%s",x);
    scanf("%s",y);
    printf("%d\n",lcs(x,y));
  }
  
  
  return 0;
}