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 long long N = 5e5 + 10;
  long long q;
  long long st[N], ed[N], last[N], ans[N], cnt;
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cin >> q;
    while (q--) {
      long long op;
      long long x;
      cin >> op >> x;
      if (op == 1) {
        if (!st[x]) {
          st[x] = ed[x] = ++cnt;
        } else {
          last[++cnt] = ed[x];
          ed[x] = cnt;
        }
      } else {
        long long y;
        cin >> y;
        if (x == y || !st[x]) continue;
        if (!st[y]) {
          st[y] = st[x];
          ed[y] = ed[x];
        } else {
          last[st[x]] = ed[y];
          ed[y] = ed[x];
        }
        st[x] = ed[x] = 0;
      }
    }
    for (long long i = (1); i <= (5e5); i++) {
      for (long long j = ed[i]; j; j = last[j]) ans[j] = i;
    }
    for (long long i = (1); i <= (cnt); i++) cout << ans[i] << " ";
  }
  