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;
  const int maxn = 5e5 + 5, INF = 0x3f3f3f3f;
  int h[maxn], t[maxn], e[maxn], ne[maxn], cur, idx;
  int ans[maxn];
  int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    memset(h, -1, sizeof h);
    memset(ne, -1, sizeof ne);
    int n;
    cin >> n;
    while (n--) {
      int op, x, y;
      cin >> op;
      if (op == 1) {
        cin >> x;
        e[idx] = ++cur;
        if (h[x] == -1) {
          h[x] = idx;
          t[idx] = idx;
        } else {
          ne[t[h[x]]] = idx;
          t[h[x]] = idx;
        }
        ++idx;
      } else {
        cin >> x >> y;
        if (x == y || h[x] == -1)
          continue;
        else if (h[y] == -1)
          h[y] = h[x];
        else {
          ne[t[h[y]]] = h[x];
          t[h[y]] = t[h[x]];
        }
        h[x] = -1;
      }
    }
    for (int x = 1; x <= 500000; x++) {
      for (int i = h[x]; ~i; i = ne[i]) ans[e[i]] = x;
    }
    for (int i = 1; i <= idx; i++) cout << ans[i] << ' ';
    cout << '\n';
  }
  