#include <iostream>
#include <string>
using namespace std;
string a,b;
int la,lb;
int dp[10000][10000];

void dpf(void){
	for(int i=1;i<=la;i++){
		for(int j=1;j<=lb;j++){
			if(a[i]==b[j]){
				dp[i][j]=dp[i-1][j-1]+1;
			}
			else{
				dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
			}
		}
	}	
	return ;
}
int n;

int main(void){
	cin>>n;
	
	
for(int i=0;i<n;i++){	
	cin>>a>>b;
	la=a.length();
	lb=b.length();
	a=' '+a;
	b=' '+b;
	dpf();
	cout<<dp[la][lb]<<endl;
}


}

### Prompt

Create a solution in CPP for the following problem:

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.

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

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.

Example

Input

3
abcbdab
bdcaba
abc
abc
abc
bc


Output

4
3
2

### Response

```cpp
#include <iostream>
#include <string>
using namespace std;
string a,b;
int la,lb;
int dp[10000][10000];

void dpf(void){
	for(int i=1;i<=la;i++){
		for(int j=1;j<=lb;j++){
			if(a[i]==b[j]){
				dp[i][j]=dp[i-1][j-1]+1;
			}
			else{
				dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
			}
		}
	}	
	return ;
}
int n;

int main(void){
	cin>>n;
	
	
for(int i=0;i<n;i++){	
	cin>>a>>b;
	la=a.length();
	lb=b.length();
	a=' '+a;
	b=' '+b;
	dpf();
	cout<<dp[la][lb]<<endl;
}


}

```