description: You have an array of integers (initially empty).
  
  You have to perform q queries. Each query is of one of two types: 
  
    * "1 x" — add the element x to the end of the array; 
    * "2 x y" — replace all occurrences of x in the array with y. 
  
  
  
  Find the resulting array after performing all the queries.
  
  Input
  
  The first line contains a single integer q (1 ≤ q ≤ 5 ⋅ 10^5) — the number of queries.
  
  Next q lines contain queries (one per line). Each query is of one of two types: 
  
    * "1 x" (1 ≤ x ≤ 5 ⋅ 10^5); 
    * "2 x y" (1 ≤ x, y ≤ 5 ⋅ 10^5). 
  
  
  
  It's guaranteed that there is at least one query of the first type.
  
  Output
  
  In a single line, print k integers — the resulting array after performing all the queries, where k is the number of queries of the first type.
  
  Examples
  
  Input
  
  
  7
  1 3
  1 1
  2 1 2
  1 2
  1 1
  1 2
  2 1 3
  
  
  Output
  
  
  3 2 2 3 2 
  
  Input
  
  
  4
  1 1
  1 2
  1 1
  2 2 2
  
  
  Output
  
  
  1 2 1 
  
  Input
  
  
  8
  2 1 4
  1 1
  1 4
  1 2
  2 2 4
  2 4 3
  1 2
  2 2 7
  
  
  Output
  
  
  1 3 3 7 
  
  Note
  
  In the first example, the array changes as follows:
  
  [] → [3] → [3, 1] → [3, 2] → [3, 2, 2] → [3, 2, 2, 1] → [3, 2, 2, 1, 2] → [3, 2, 2, 3, 2].
  
  In the second example, the array changes as follows:
  
  [] → [1] → [1, 2] → [1, 2, 1] → [1, 2, 1].
  
  In the third example, the array changes as follows:
  
  [] → [] → [1] → [1, 4] → [1, 4, 2] → [1, 4, 4] → [1, 3, 3] → [1, 3, 3, 2] → [1, 3, 3, 7].
solution: import java.io.*;
  import java.util.*;
  import static java.lang.Math.*;
  
  public class E{
  	
  	static List<Integer>[] a;
  	static int n;
  	static int[] ptr;
  	
  	static void merge(int x, int replace) {
  		if (x==replace) return;
  		if (a[ptr[replace]].size() > a[ptr[x]].size()) {
  			// good
  		}
  		else {
  			ptr[replace]^=ptr[x];
  			ptr[x]^=ptr[replace];
  			ptr[replace]^=ptr[x];
  		}
  		a[ptr[replace]].addAll(a[ptr[x]]);
  		a[ptr[x]].clear();
  		
  	}
  	
  	public static void main(String[] args) throws IOException {
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		//new Thread(null, new (), "peepee", 1<<28).start();
  		
  		n =readInt();
  		a = new List[(int)5e5+1]; ptr = new int[a.length];
  		for (int i =0 ; i < a.length; i++) a[ptr[i] = i] = new ArrayList<Integer>();
  		int shit = 0;
  		while(n-->0) {
  			int t= readInt();
  			if (t==1) {
  				int x =readInt();
  				a[ptr[x]].add(shit++);
  			}
  			else {
  				int x = readInt();
  				int y =readInt();
  				merge(x,y);
  			}
  		}
  		int[] ans = new int[shit];
  		for (int i = 0; i < a.length; i++) {
  			for (int x: a[ptr[i]]) {
  				ans[x] = i;
  			}
  		}
  		for (int x: ans) out.print(x + " ");
  		out.println();
  		out.close();
  	}
  	
  	/* Stupid things to try if stuck:
  	 * n=1, expand bsearch range
  	 * brute force small patterns
  	 * submit stupid intuition
  	 */
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
  	static int readInt() throws IOException{return Integer.parseInt(read());}
  	static long readLong() throws IOException{return Long.parseLong(read());}
  	static double readDouble() throws IOException{return Double.parseDouble(read());}
  }