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>
using namespace std;

int main() {
  int n, num_query; cin >> n >> num_query;
  queue<int> que[num_query];
  
  for (int loop = 0; loop < num_query; loop++) {
    int command, a, b; scanf("%d %d", &command, &a);
    
    switch (command) {
      case 0:
        scanf("%d", &b);
        que[a].push(b);
        break;
        
      case 1:
        if (!que[a].empty()) printf("%d\n", que[a].front());
        break;
        
      case 2:
        if (!que[a].empty()) que[a].pop();
        break;
    }
  }
}