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 <bits/stdc++.h>

using namespace std;
#define rep_i(i, n) for (int i = 0; i < (n); ++i)
#define rep_ll(i, n) for (long long i = 0; i < (n); ++i)
#define r_rep_i(i, start, end) for (int i = (start); i < (end); ++i)
#define r_rep_ll(i, start, end) for (long long i = (start); i < (end); ++i)
#define debug_vi(v) copy((v).begin(), (v).end(), ostream_iterator<int>(cout, " "));
#define debug_vll(v) copy((v).begin(), (v).end(), ostream_iterator<long long>(cout, " "));
#define debug_vd(v) copy((v).begin(), (v).end(), ostream_iterator<double>(cout, " "));
#define sort_v(v) sort((v).begin(), (v).end()); //昇順
#define d_sort_vi(v) sort((v).begin(), (v).end(), greater<int>()); //降
#define d_sort_vd(v) sort((v).begin(), (v).end(), greater<double>()); //昇順
#define say(t) cout << (t) << endl;
#define sum_vi(v) accumulate((v).begin(), (v).end(), 0);
#define sum_vll(v)accumulate((v).begin(), (v).end(), 0LL);

int main(){
  int n, q;
  cin >> n >> q;
  vector< queue<int> > v(n);
  rep_i(i, n){
    queue<int> a;
    v.push_back(a);
  }

  int tmp, tmp2, tmp3;
  rep_i(i, q){
    cin >> tmp;
    switch (tmp)
    {
    case 0:
      cin >> tmp2 >> tmp3;
      v[tmp2].push(tmp3);
      break;
    case 1:
      cin >> tmp2;
      if(v[tmp2].empty()){
        break;
      }
      say(v[tmp2].front());
      break;
    case 2:
      cin >> tmp2;
      if(v[tmp2].empty()){
        break;
      }
      v[tmp2].pop();
      break;
    default:
      break;
    }
  }
}