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

Solution (in CPP):

Solution:
#include <iostream>
#include <queue>
#include <stdio.h>
#include <vector>
using namespace std;

int main()
{
    int n, q;
    scanf("%d %d", &n, &q);
    vector<queue<int> > vec(n);

    int flag, t, x;
    for (int i = 0; i < q; i++)
    {
        scanf("%d", &flag);
        switch (flag) {
        case 0:
            scanf("%d %d", &t, &x);
            vec[t].push(x);
            break;
        case 1:
            scanf("%d", &t);
            if (!vec[t].empty())
            {
                printf("%d\n",vec[t].front());
            }
            break;
        case 2:
            scanf("%d", &t);
            if (!vec[t].empty())
            {
                vec[t].pop();
            }
            break;
        default:
            break;
        }
    }


    return 0;
}