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>

using namespace std;
#define MAX 1000

class queue{
    int num;
    queue* head = NULL;
    queue* tail = NULL;
public:
    void enque(int x){
        if(head == NULL){
            head = new queue(x);
            tail = head;
        }
        else{
            tail->head = new queue(x);
            tail = tail->head;
        }
    }
    
    int front(){
        return head->num;
    }
    
    void deque(){
        // 繋ぎかえる
        if(this->head->head != NULL){
            queue* tmp = head;
            this->head = this->head->head;
            delete tmp;
            tmp = NULL;
        }
        else if(head != NULL){
            delete head;
            head = NULL;
        }
    }
    
    bool empty(){
        return head == NULL;
    }
    
    queue(){
        
    }
    
    queue(int num){
        this->num = num;
    }
};

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n,q,command,t,x;
    queue myQueue[MAX];
    cin >> n >> q;
    
    for (int i = 0;i < q;++i) {
        cin >> command >> t;
        switch(command){
                // enque
            case 0:
                cin >> x;
                myQueue[t].enque(x);
                break;
                // front
            case 1:
                if(!myQueue[t].empty()){
                    cout << myQueue[t].front() << endl;
                }
                break;
                // deque
            case 2:
                if(!myQueue[t].empty()){
                    myQueue[t].deque();
                }
                break;
        }
    }
}