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 io, os
  input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  from collections import defaultdict
  
  def main():
  
      q = int(input())
      cur = 0
      D = defaultdict(lambda: [])
      for i in range(q):
          temp = list(map(int, input().split()))
          if temp[0] == 1:
              x = temp[1]
              D[x].append(cur)
              cur += 1
          else:
              x, y = temp[1:]
              if x == y:
                  continue
              lx = D[x]
              ly = D[y]
              if len(ly) >= len(lx):
                  for j in lx:
                      D[y].append(j)
                  D.pop(x)
              else:
                  for j in ly:
                      D[x].append(j)
                  D[y] = D[x]
                  D.pop(x)
      #print(D)
      ans = [0]*cur
      for k, l in D.items():
          for i in l:
              ans[i] = k
      print(*ans)
  
  if __name__ == '__main__':
      main()
  