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 sys
  import math
  from collections import deque,Counter
  from sys import stdin
  
  #sys.setrecursionlimit(10**7)
  
  int1=lambda x: int(x)-1
  stinput=lambda :stdin.readline()[:-1]
  ii=lambda :int(stinput())
  mi=lambda :map(int, stdin.readline().split())
  li=lambda :list(mi())
  mi1=lambda :map(int1, stdin.readline().split())
  li1=lambda :list(mi1())
  mis=lambda :map(str, stdin.readline().split())
  lis=lambda :list(mis())
  
  from collections import defaultdict
  #d=defaultdict(int)
  #d=defaultdict(lambda:1)
  
  import heapq
  #heapq.heapify(hq)
  #heapq.heappop(hq)
  #heapq.heappush(hq,1)
  
  mod=10**9+7
  Mod=998244353
  INF=10**18
  ans=0
  
  q=ii()
  query=[]
  for i in range(q):
    query.append(li())
  
  ans=[]
  d=defaultdict(int)
  for x in query[::-1]:
    if x[0]==2:
      y,z=x[1],x[2]
      if z in d:
        d[y]=d[z]
      else:
        d[y]=z
    else:
      y=x[1]
      if y in d:
        ans.append(d[y])
      else:
        ans.append(y)
  print(*ans[::-1])