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 <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dy[] = {1, 0, -1, 0};
const int dx[] = {0, 1, 0, -1};
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, q;
    cin >> n >> q;
    queue<int> que[n];
    for(int i = 0; i < q; i++)
    {
        int v, t, x;
        cin >> v;
        if(v == 0)
        {
            cin >> t >> x;
            que[t].push(x);
        }
        else if(v == 1)
        {
            cin >> t;
            if(que[t].size() != 0) cout << que[t].front() << "\n";
        }
        else if(v == 2)
        {
            cin >> t;
            if(que[t].size() != 0) que[t].pop();
        }
    }
    return 0;
}