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 collections import defaultdict as dd
  import math
  import sys
  input=sys.stdin.readline
  def nn():
      return int(input())
   
  def li():
      return list(input())
   
  def mi():
      return map(int, input().split())
   
  def lm():
      return list(map(int, input().split()))
   
   
   
  def solve():
      n = nn()
      
      dlists = dd(list)
      dkeys = dict()
      place = 0
      id_vals = 10**6
      for i in range(n):
          q = lm()
          if q[0] == 1:
              if q[1] in dkeys:
                  val = dkeys[q[1]]
                  dlists[val].append(place)
                  
              else:
                  dkeys[q[1]] = id_vals
                  dlists[id_vals].append(place)
                  
                  id_vals+=1
              place +=1
          else:
              
              x,y = q[1],q[2]
              
              if x==y:
                  pass
              elif x in dkeys:
                  if y in dkeys:
                      yid = dkeys[y]
                      xid = dkeys[x]
                      
                      ylen = len(dlists[yid])
                      xlen = len(dlists[xid])
                      if xlen<=ylen:
                          dlists[yid].extend(dlists[xid])
                          dkeys[y] = yid
                      else:
                          dlists[xid].extend(dlists[yid])
                          dkeys[y] = xid
                          
                      del dkeys[x]
                      
                  else:
                      dkeys[y] = dkeys[x]
                      del dkeys[x]
                  
          #print(dlists)
          #print(dkeys)
              
      out = [0 for i in range(place)]
      
      for key,value in dkeys.items():
          nums = dlists[value]
          for num in nums:
              out[num] = key
      #print(dlists)
      #print(dkeys)
      print(*out)
      
      
              
                  
      
  
  
  q=1
  for _ in range(q):
      solve()