#include <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;

const int MOD_TYPE = 1;
const ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
const int INF = (int)1e9;
const ll LINF = (ll)4e18;
const double PI = acos(-1.0);

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define possible(n) cout << ((n) ? "possible" : "impossible") << endl
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << endl;

vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(50) << setiosflags(ios::fixed);
  int n, q;
  cin >> n >> q;
  list<int> ls[1000];
  rep(qi, q)
  {
    int type;
    cin >> type;
    if (type == 0)
    {
      int t, x;
      cin >> t >> x;
      ls[t].push_back(x);
    }
    else if (type == 1)
    {
      int t;
      cin >> t;
      for (auto itr = ls[t].begin(); itr != ls[t].end(); itr++)
        cout << (itr == ls[t].begin() ? "" : " ") << *itr;
      cout << endl;
    }
    else
    {
      int s, t;
      cin >> s >> t;
      ls[t].splice(ls[t].end(), ls[s]);
    }
  }

  return 0;
}


### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;

const int MOD_TYPE = 1;
const ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
const int INF = (int)1e9;
const ll LINF = (ll)4e18;
const double PI = acos(-1.0);

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define possible(n) cout << ((n) ? "possible" : "impossible") << endl
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << endl;

vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(50) << setiosflags(ios::fixed);
  int n, q;
  cin >> n >> q;
  list<int> ls[1000];
  rep(qi, q)
  {
    int type;
    cin >> type;
    if (type == 0)
    {
      int t, x;
      cin >> t >> x;
      ls[t].push_back(x);
    }
    else if (type == 1)
    {
      int t;
      cin >> t;
      for (auto itr = ls[t].begin(); itr != ls[t].end(); itr++)
        cout << (itr == ls[t].begin() ? "" : " ") << *itr;
      cout << endl;
    }
    else
    {
      int s, t;
      cin >> s >> t;
      ls[t].splice(ls[t].end(), ls[s]);
    }
  }

  return 0;
}


```