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<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define NUM 1000

int N,num_query;
queue<int> Q[NUM];

int main(){

	scanf("%d %d",&N,&num_query);

	int command,A,B;

	for(int loop = 0; loop < num_query; loop++){

		scanf("%d",&command);

		switch(command){
		case 0:

			scanf("%d %d",&A,&B);
			Q[A].push(B);
			break;

		case 1:

			scanf("%d",&A);
			if(!Q[A].empty()){
				printf("%d\n",Q[A].front());
			}
			break;

		case 2:

			scanf("%d",&A);
			if(!Q[A].empty()){
				Q[A].pop();
			}
			break;
		}
	}
	return 0;
}