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 <iostream>
#include <fstream>
#include <queue>
//#define LOCAL
using namespace std;
typedef long long ll;
const int mx=1e3+10;
queue<ll>que[mx];
int main() {
#ifdef LOCAL
ifstream fin("1.txt");
streambuf*p;
p=cin.rdbuf(fin.rdbuf());
#endif // LOCAL
ll n, q;
cin>>n>>q;
for(ll i=1;i<=q;i++){
    ll qn, sub, value;
 //   int sub, que, value;
    cin>>qn;
    switch(qn){
        case 0: cin>>sub>>value;
                que[sub].push(value);
                break;
        case 1: cin>>sub;
                if(!que[sub].empty()) cout<<que[sub].front()<<endl;
                break;
        case 2: cin>>sub;
                if(!que[sub].empty()) que[sub].pop();
    }
}


#ifdef LOCAL
fin.close();
#endif // LOCAL
}