#include<iostream>
#include<queue>         //FIFO=first in,first out
using namespace std;
int main()
{
   int n,q,index,order,value,x;
  cin>>n>>q;
  queue<int>Q[n];
  while(q--){


    cin>>order>>index;

  switch(order)
  {
      case 0:
                  cin>>value;
                  Q[index].push(value);
                  break;
      case 1:
                  if( !Q[index].empty())
                  cout<< Q[index].front()<<endl;
                     break;
      case 2:
                   if( !Q[index].empty())
                  Q[index].pop();
                  break;
  }

  }
return 0;
}


### Prompt

Please formulate 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<queue>         //FIFO=first in,first out
using namespace std;
int main()
{
   int n,q,index,order,value,x;
  cin>>n>>q;
  queue<int>Q[n];
  while(q--){


    cin>>order>>index;

  switch(order)
  {
      case 0:
                  cin>>value;
                  Q[index].push(value);
                  break;
      case 1:
                  if( !Q[index].empty())
                  cout<< Q[index].front()<<endl;
                     break;
      case 2:
                   if( !Q[index].empty())
                  Q[index].pop();
                  break;
  }

  }
return 0;
}


```