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 math import *
   
  #from math import factorial as fact, comb as ncr 
  from bisect import bisect_left as bl
  from bisect import bisect_right as br
  from collections import Counter as ctr
  from collections import deque as dq
  
  from array import array
  from re import search
   
  li=lambda : list(map(int,input().split()))
  arr=lambda a: array('i',a)
  bi=lambda n: bin(n).replace("0b", "")
  yn=lambda f: print('NYOE S'[f::2])
  sbstr=lambda a,s: search('.*'.join(a),s)
  
  '''def isprime(a):
      if a==2: return True
      elif a%2==0: return False
      n=3
      while n<=a**0.5:
          if a%n==0:
              return False
          n+=2
      return True
  '''
  '''def nf(n,a):
      b=a*a
      c=b*a
      d=c*a
      e=n
      ct=0
      while e%a==0:
          if e%d==0:
              ct+=4
              e=e//d
          if e%c==0:
              ct+=3
              e=e//c
          if e%b==0:
              ct+=2
              e=e//b
          if e%a==0:
              ct+=1
              e=e//a
      return ct
  '''
  
  def solve():
      ans=[]
      qre=int(input())
      q=[]
      for _ in range(qre):
          q.append(li())
      #print(q)
      sw=dict()
      for qr in reversed(q):
         # print(qr)
          if qr[0]==1:
              if qr[1] in sw:
                  ans.append(sw[qr[1]])
              else:
                  ans.append(qr[1])
          else:
              if qr[2] not in sw:
                  sw[qr[1]]=qr[2]
              else:
                  sw[qr[1]]=sw[qr[2]]
          #print(*ans)
      
      
      print(*ans[::-1])
  
  solve()