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 JAVA):

Solution:
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		List<Deque<Integer>> list = new ArrayList<>();
		int n = sc.nextInt();
		int q = sc.nextInt();
		for (int i = 0; i < n; i++) {
			list.add(new ArrayDeque<>());
		}
		for (int i = 0; i < q; i++) {
			switch (sc.nextInt()) {
			case 0:
				list.get(sc.nextInt()).addLast(sc.nextInt());
				break;
			case 1:
				int t = sc.nextInt();
				if (!list.get(t).isEmpty()) {
					System.out.println(list.get(t).peekFirst());
				}
				break;
			case 2:
				t = sc.nextInt();
				if (!list.get(t).isEmpty()) {
					list.get(t).removeFirst();
				}
				break;
			}
		}
		sc.close();
	}
}