#include <iostream>
using namespace std;

void solve(){
    string S, T;
    cin >> S;
    cin >> T;

    int dp[1001][1001];
    dp[0][0] = 0;

    for (int i = 0; i < S.size(); i++){
        for (int j = 0; j < T.size(); j++){
            if (S[i] == T[j])
                dp[i+1][j+1] = dp[i][j] + 1;
            else
                dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
        }
    }
    cout << dp[S.size()][T.size()] << endl;
}

int main(){
    int q;
    cin >> q;

    for (int i = 0; i < q; i++)
        solve();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;

void solve(){
    string S, T;
    cin >> S;
    cin >> T;

    int dp[1001][1001];
    dp[0][0] = 0;

    for (int i = 0; i < S.size(); i++){
        for (int j = 0; j < T.size(); j++){
            if (S[i] == T[j])
                dp[i+1][j+1] = dp[i][j] + 1;
            else
                dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
        }
    }
    cout << dp[S.size()][T.size()] << endl;
}

int main(){
    int q;
    cin >> q;

    for (int i = 0; i < q; i++)
        solve();
}

```