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 n;
  struct Node {
    int val, nxt;
  } node[500005];
  int tot = 0;
  int st[500005];
  int ed[500005];
  int ans[500005];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      int op, x, y;
      cin >> op >> x;
      if (op == 1) {
        node[++tot].val = ++n;
        if (!st[x]) {
          st[x] = ed[x] = tot;
          continue;
        }
        node[tot].nxt = st[x];
        st[x] = tot;
      } else {
        cin >> y;
        if (y == x) continue;
        if (!st[x]) continue;
        if (!st[y]) {
          st[y] = st[x];
          ed[y] = ed[x];
          st[x] = ed[x] = 0;
          continue;
        }
        node[ed[y]].nxt = st[x];
        ed[y] = ed[x];
        st[x] = ed[x] = 0;
      }
    }
    for (int i = 1; i < 500005; i++) {
      int now = st[i];
      while (now) {
        ans[node[now].val] = i;
        now = node[now].nxt;
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << ' ';
    }
    return 0;
  }
  