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: from sys import stdin,stdout,setrecursionlimit
  from math import gcd,sqrt,factorial,pi,inf
  from collections import deque,defaultdict
  from bisect import bisect,bisect_left
  from time import time
  from itertools import permutations as per
  from heapq import heapify,heappush,heappop,heappushpop
  input=stdin.readline
  R=lambda:map(int,input().split())
  I=lambda:int(input())
  S=lambda:input().rstrip('\r\n')
  L=lambda:list(R())
  P=lambda x:stdout.write(str(x)+'\n')
  lcm=lambda x,y:(x*y)//gcd(x,y)
  nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
  inv=lambda x:pow(x,N-2,N)
  sumx=lambda x:(x**2+x)//2
  N=10**9+7
  
  a=[]
  p=[i for i in range(5*(10**5)+1)]
  for i in [L() for i in range(I())][::-1]:
  	if i[0]==1:
  		a+=p[i[1]],
  	else:
  		p[i[1]]=p[i[2]]
  print(*a[::-1])