#include<iostream>
#include<string>
#include<math.h>
#include<algorithm>
#include<vector>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
int main(){
    int n,dp[1001][1001];
    cin>>n;
    string x,y;
    rep(i,n){
        cin>>x>>y;
        rep(j,x.size()+1) rep(k,y.size()+1) dp[j][k]=0;
        rep(j,x.size()) rep(k,y.size()){
            if(x[j]==y[k]) dp[j+1][k+1]=max(max(dp[j][k]+1,dp[j+1][k]),dp[j][k+1]);
            else dp[j+1][k+1]=max(dp[j+1][k],dp[j][k+1]);
        }
        cout<<dp[x.size()][y.size()]<<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>
#include<math.h>
#include<algorithm>
#include<vector>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
int main(){
    int n,dp[1001][1001];
    cin>>n;
    string x,y;
    rep(i,n){
        cin>>x>>y;
        rep(j,x.size()+1) rep(k,y.size()+1) dp[j][k]=0;
        rep(j,x.size()) rep(k,y.size()){
            if(x[j]==y[k]) dp[j+1][k+1]=max(max(dp[j][k]+1,dp[j+1][k]),dp[j][k+1]);
            else dp[j+1][k+1]=max(dp[j+1][k],dp[j][k+1]);
        }
        cout<<dp[x.size()][y.size()]<<endl;
    }
}

```