#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
int n, Q;
queue<int>q[1010];
int main() {
	cin >> n >> Q;
	while (Q--) {
		int z, t, x;
		cin >> z >> t;
		if (z == 0) {
			cin >> x;
			q[t].push(x);
		}
		if (z == 1 && !q[t].empty())cout << q[t].front() << endl;
		if (z == 2 && !q[t].empty())q[t].pop();
	}
	getchar();getchar();
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
int n, Q;
queue<int>q[1010];
int main() {
	cin >> n >> Q;
	while (Q--) {
		int z, t, x;
		cin >> z >> t;
		if (z == 0) {
			cin >> x;
			q[t].push(x);
		}
		if (z == 1 && !q[t].empty())cout << q[t].front() << endl;
		if (z == 2 && !q[t].empty())q[t].pop();
	}
	getchar();getchar();
	return 0;
}

```