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: #include <bits/stdc++.h>
  using namespace std;
  int q;
  const int N = 500090;
  int pa[N], co[N], ch[N], val[N];
  int get(int x) {
    if (pa[x] == x) return x;
    return pa[x] = get(pa[x]);
  }
  int main() {
    cin >> q;
    int n = 0;
    memset(ch, -1, sizeof ch);
    memset(co, -1, sizeof co);
    for (int i = 0; i < q; i++) {
      int t, x, y;
      cin >> t >> x;
      if (t == 1) {
        if (ch[x] == -1) ch[x] = i;
        pa[i] = ch[x];
        co[i] = x;
        n++;
        val[i] = 1;
        continue;
      }
      cin >> y;
      int z = ch[x];
      if (z == -1) continue;
      ch[x] = -1;
      co[z] = y;
      if (ch[y] != -1) pa[ch[y]] = z;
      ch[y] = z;
    }
    for (int i = 0; i < q; i++) {
      if (val[i] == 0) continue;
      cout << co[get(i)] << " ";
    }
  }
  