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 <iostream>
#include <iomanip>
#include<vector>
#include <algorithm>
#include <queue>
#include<string>
#include <map>
#include <cmath>
#include <deque>
#include <list>
#include <stack>
#include <queue>
#include <utility>
typedef long long ll;
typedef long double ld;

#define REP(i,s,n)for(ll i=s;i<(n);++i)
#define rep(i,n) REP(i,0,n)
using namespace std;
int S[1000] = { 0 };


int main() {
	ll N,Q;
	cin >>N>>Q;
	list<ll>Y[N];
	rep(i, Q) {
		int a;
		cin >> a;
		if (a == 0) {
			int b,c;
			cin >> b>>c;
			Y[b].push_back(c);
		}
		else if (a == 1) {
			int b;
			cin >> b;
			if (0<Y[b].size()) {
				list<ll>::iterator flg = Y[b].begin();
				int counter = Y[b].size()-1;
				for (; flg != Y[b].end(); ++flg) {
					if (0 < counter) {
						cout << *flg <<" ";
						counter--;
					}
					else {
						break;
					}
				}
				cout << *flg << endl;
			}
			else {
				cout << endl;
			}
		}
		else {
			int b,c;
			cin >> b>>c;
			Y[c].splice(Y[c].end(), Y[b]);
		}
	}
	return 0;
}