#include<bits/stdc++.h>
using namespace std;
 
 int c[4001][4001]{};
signed main(){
    int n;
    scanf("%d",&n);
    char a[5000];
char b[5000];
    for(int m=0;m<n;m++){
	scanf("%s%s",a,b);
int d,e;
d=strlen(a);
e=strlen(b);
for(int t=0;t<=d;t++){
    for(int w=0;w<=d;w++)c[t][w]=0;
    }
for(int x=0;x<d;x++){
    for(int y=0;y<e;y++){
	if(a[x]==b[y])c[x+1][y+1]=c[x][y]+1;
	else c[x+1][y+1]=max(c[x][y+1],c[x+1][y]);
	}
    }
    printf("%d\n",c[d][e]);
	}
}
### Prompt

Your task is to create a Cpp solution to 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<bits/stdc++.h>
using namespace std;
 
 int c[4001][4001]{};
signed main(){
    int n;
    scanf("%d",&n);
    char a[5000];
char b[5000];
    for(int m=0;m<n;m++){
	scanf("%s%s",a,b);
int d,e;
d=strlen(a);
e=strlen(b);
for(int t=0;t<=d;t++){
    for(int w=0;w<=d;w++)c[t][w]=0;
    }
for(int x=0;x<d;x++){
    for(int y=0;y<e;y++){
	if(a[x]==b[y])c[x+1][y+1]=c[x][y]+1;
	else c[x+1][y+1]=max(c[x][y+1],c[x+1][y]);
	}
    }
    printf("%d\n",c[d][e]);
	}
}
```