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 <queue>
#include <vector>
using namespace std;
int main()
{
    typedef queue<int> ls;

    int n;
    cin >> n;

    vector<ls> Q(n);

    int q;
    cin >> q;

    while (q)
    {
        int query, t;
        cin >> query >> t;
        ls &l = Q.at(t);
        if (query == 0)
        {
            //整数xを今の末尾に代入
            int x;
            cin >> x;
            l.push(x);
        }
        if (query == 1)
        {
            //次に取り出される要素(First In First Out)つまり先頭を報告
            int begin = l.front();
            //空なら何もしない
            //empty()は？
            if (!l.empty())
            {
                cout << begin << endl;
            }
        }
        if (query == 2)
        {
            //取り出された要素を削除つまり先頭のイテレーターを末尾ので方へずらす
            //先頭の要素を削除
            // int &begin = l.front();
            //空なら何もしない
            if (!l.empty())
            {
                // ls::iterator it = l.begin(); //先頭の要素を指す
                // l.erase(it);
                l.pop();
            }
        }
        q--;
    }
}