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
  input = sys.stdin.readline
  
  class Node():
      def __init__(self,val):
          self.val = val
          self.next = None
  
  n = int(input())
  cnt = 0
  d = {}
  g = {}
  for i in range(n):
      l = list(map(int,input().split()))
      if len(l) == 2:
          x = l[1]
          now = Node(cnt)
          if x in d and d[x] != None:
              d[x].next = now
              d[x] = d[x].next
              
          else:
              d[x] = now
              g[x] = now
              
          cnt += 1
          
      else:
          x,y = l[1],l[2]
          if x not in d or d[x] == None:
              continue
          
          if x == y:
              continue
          
          if y in d and d[y] != None:
              d[y].next = g[x]
              d[y] = d[x]
              
          else:
              d[y] = d[x]
              g[y] = g[x]
              
          d[x] = None
          g[x] = None
              
  ans = [0 for i in range(cnt)]
  for x in g:
      while g[x] != None:
          ans[g[x].val] = x
          g[x] = g[x].next
          
  print(*ans)