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;
using ll = long long; using ull = unsigned long long;
#define int ll
using vb = vector<bool>; using vvb = vector<vb>;
using vi = vector<int>; using vvi = vector<vi>;
using vl = vector<ll>; using vvl = vector<vl>;
using pii = pair<int,int>; using pll = pair<ll,ll>;
#define FOR(i,a,b) for(ll i = (a); i < (ll)(b); ++i)
#define REP(i,n) FOR(i,0,n)
#define RFOR(i,a,b) for(ll i = (ll)(b)-1;i >= a;--i)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define eb(val) emplace_back(val)
const double PI = acos(-1);
const double EPS = 1e-10;
const ll MOD = 1e9+7;
void cioacc(){//accelerate cin/cout
  cin.tie(0);
  ios::sync_with_stdio(false);
}

signed main(){
  int n;
  cin >> n;
  vector<queue<int>> a(n);
  int q;
  cin >> q;
  REP(i,q){
    int b,t;
    cin >> b >> t;
    if(b==0){
      int x;
      cin >> x;
      a[t].push(x);
    }else if(b==1){
      if(!a[t].empty()){
        cout << a[t].front() << endl;
      }
    }else{
      if(!a[t].empty()){
        a[t].pop();
      }
    }
  }
}