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
  
  q = int(sys.stdin.readline())
  
  f = list(range(500001))
  dd = [0] * 500001
  
  
  def find(i):
      if i != f[i]:
          f[i] = find(f[i])
      return f[i]
  
  
  def union(i, j):
      fi, fj = find(i), find(j)
      if fi != fj:
          if dd[fi] < dd[fj]:
              f[fi] = fj
          else:
              f[fj] = fi
          if dd[fi] == dd[fj]:
              dd[fi] += 1
  
  
  mp = {}
  
  n = 0
  
  arr = []
  
  for _ in range(q):
      op = list(map(int, sys.stdin.readline().split()))
      if op[0] == 1:
          x = op[1]
          arr.append(x)
          if x in mp:
              union(mp[x], n)
          else:
              mp[x] = n
          n += 1
      else:
          x, y = op[1], op[2]
          if x == y or x not in mp:
              continue
          if y in mp:
              arr[find(mp[x])] = y
              union(mp[x], mp[y])
          else:
              arr[find(mp[x])] = y
              mp[y] = mp[x]
          del mp[x]
  for i in range(n):
      arr[i] = arr[find(i)]
  
  print(' '.join(map(str, arr)))
  