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 <string>
#include <iostream>
#include <stdio.h>
#include <queue>

using namespace std;

int main() {
  ios::sync_with_stdio(false);

  int num_queue, num_query;
  int cmd, index, val;

  cin >> num_queue >> num_query;
  queue<int> q[num_queue];

  for (int i = 0; i < num_query; i++) {
    cin >> cmd;

    switch (cmd) {
      // enqueue
      case 0:
        cin >> index >> val;
        q[index].push(val);

        break;

      case 1:
        cin >> index;
        if (!q[index].empty()) {
          cout << q[index].front() << endl;
        }

        break;

      case 2:
        cin >> index;
        if (!q[index].empty()) {
          q[index].pop();
        }
        break;
    }
  }
}