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<stdlib.h>
 
static const int N = 1000;
 
int max(int s,int t){
if (s>=t) return s;
return t;
}
 
int lcs(char X[], char Y[], int m ,int n) {
  int c[N + 1][N +1];
  int maxl = 0,i,j;
 
  strcmp(&X[0], " ");
  strcmp(&Y[0], " ");
  //X[0] = ' ';
  //Y[0] = ' ';
  for(i=0;i<=m;i++) c[i][0]=0;
  for(j=0;j<=n;j++) c[0][j]=0;
 
  for(i = 1;i <=m; i++) {
    for(j = 1;j <=n; j++) {
    if(X[i] == Y[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() {
  char s1[1001],s2[1001];
  int i,n;
 
  scanf("%d\n", &n);
  for (i=0;i<n;i++) {
    scanf("%s", &s1[1]);
    scanf("%s", &s2[1]);
    printf("%d\n", lcs(s1,s2,(int)strlen(&s1[1]),(int)strlen(&s2[1])));
  }
  return 0;
}