#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
const ll inf = 1e9;
int main(){
    int n, q;
    cin >> n >> q;
    vector<queue<ll> > vs(n);
    rep(i,q){
        ll com, t; 
        cin >> com >> t;
        ll x;
        if(com == 0){
            cin >> x;
            vs[t].push(x);
        }else if(com == 1 && vs[t].size() > 0)cout << vs[t].front() << endl;
        else if(vs[t].size() > 0)vs[t].pop();
    }
    return 0;
} 

### Prompt

Please create a solution in cpp to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
const ll inf = 1e9;
int main(){
    int n, q;
    cin >> n >> q;
    vector<queue<ll> > vs(n);
    rep(i,q){
        ll com, t; 
        cin >> com >> t;
        ll x;
        if(com == 0){
            cin >> x;
            vs[t].push(x);
        }else if(com == 1 && vs[t].size() > 0)cout << vs[t].front() << endl;
        else if(vs[t].size() > 0)vs[t].pop();
    }
    return 0;
} 

```