Problem: For $n$ lists $L_i$ $(i = 0, 1, ..., n-1)$, perform a sequence of the following operations.

* insert($t$, $x$): Insert an integer $x$ at the end of $L_t$.
* dump($t$): Print all elements in $L_t$.
* splice($s$, $t$): Transfer elements of $L_s$ to the end of $L_t$. $L_s$ becomes empty.



In the initial state, $L_i$ $(i = 0, 1, ..., n-1)$ are empty.

Constraints

* $1 \leq n \leq 1,000$
* $1 \leq q \leq 500,000$
* For a splice operation, $s \ne t$
* For a splice operation, $L_s$ is not empty
* The total number of elements printed by dump operations do not exceed 1,000,000
* $-1,000,000,000 \leq x \leq 1,000,000,000$

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 $s$ $t$


where the first digits 0, 1 and 2 represent insert, dump and splice operations respectively.

Output

For each dump operation, print elements of the corresponding list in a line. Separete adjacency elements by a space character (do not print the space after the last element). Note that, if the list is empty, an empty line should be printed.

Example

Input

3 10
0 0 1
0 0 2
0 0 3
0 1 4
0 1 5
2 1 0
0 2 6
1 0
1 1
1 2


Output

1 2 3 4 5

6

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>;
template<class T> using V = vector<T>;
template<class T> using VV = vector<V<T>>;
#define fi first
#define se second
#define maxs(x,y) (x=max(x,y))
#define mins(x,y) (x=min(x,y))
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), EPS = 1e-10;
const ll MOD = 1e9+7;
void cioacc(){//accelerate cin/cout
  cin.tie(0);
  ios::sync_with_stdio(false);
}
void printvec(list<int> a){
  if(a.empty()){
    cout << endl;
    return;
  }
  for(auto itr=a.begin();itr!=a.end();++itr){
    auto itr2 = itr;
    itr2++;
    if(itr2==a.end()) cout << *itr << endl;
    else cout << *itr << " ";
  }
}
signed main(){
  int n;
  cin >> n;
  vector<list<int>> v(n);
  int q;
  cin >> q;
  REP(i,q){
    int d,t;
    cin >> d >> t;
    if(d==0){
      int x;
      cin >> x;
      v[t].eb(x);
    }else if(d==1){
      printvec(v[t]);
    }else{
      int s;
      cin >> s;
      v[s].splice(v[s].end(),v[t]);
      v[t].clear();
    }
  }
}