#include <iostream>
#include <iterator>
#include <queue>
#include <vector>

using namespace std;

int main(){
    int n, q;
    cin >> n >> q;
    vector<queue<int>> A(n);
    int x, y, z;
    for(int i = 0; i < q; i++){
        cin >> x;
        if(x == 0){
            cin >> y >> z;
            A[y].push(z);
        }else if(x == 1){
            cin >> y;
            if(A[y].empty()){
                continue;
            }else{
                cout << A[y].front() << "\n";
            }
        }else{
            cin >> y;
            if(A[y].empty()){
                continue;
            }else{
                A[y].pop();
            }
        }
    }

    return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

Queue is a container of elements that are inserted and deleted according to FIFO (First In First Out).

For $n$ queues $Q_i$ ($i = 0, 1, ..., n-1$), perform a sequence of the following operations.

* enqueue($t$, $x$): Insert an integer $x$ to $Q_t$.
* front($t$): Report the value which should be deleted next from $Q_t$. If $Q_t$ is empty, do nothing.
* dequeue($t$): Delete an element from $Q_t$. If $Q_t$ is empty, do nothing.



In the initial state, all queues are empty.

Constraints

* $1 \leq n \leq 1,000$
* $1 \leq q \leq 200,000$
* $-1,000,000,000 \leq x \leq 1,000,000,000$

Input

The input is given in the following format.


$n \; q$
$query_1$
$query_2$
:
$query_q$


Each query $query_i$ is given by


0 $t$ $x$


or


1 $t$


or


2 $t$


where the first digits 0, 1 and 2 represent enqueue, front and dequeue operations respectively.

Output

For each front operation, print an integer in a line.

Example

Input

3 9
0 0 1
0 0 2
0 0 3
0 2 4
0 2 5
1 0
1 2
2 0
1 0


Output

1
4
2

### Response

```cpp
#include <iostream>
#include <iterator>
#include <queue>
#include <vector>

using namespace std;

int main(){
    int n, q;
    cin >> n >> q;
    vector<queue<int>> A(n);
    int x, y, z;
    for(int i = 0; i < q; i++){
        cin >> x;
        if(x == 0){
            cin >> y >> z;
            A[y].push(z);
        }else if(x == 1){
            cin >> y;
            if(A[y].empty()){
                continue;
            }else{
                cout << A[y].front() << "\n";
            }
        }else{
            cin >> y;
            if(A[y].empty()){
                continue;
            }else{
                A[y].pop();
            }
        }
    }

    return 0;
}

```