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 = 500100;
  void solve() {
    long long q;
    cin >> q;
    vector<pair<long long, long long>> v(q);
    vector<long long> par(N), ans;
    for (long long i = 0; i < q; i++) {
      long long type;
      cin >> type;
      if (type == 1) {
        long long x;
        cin >> x;
        v[i] = {x, -1};
      } else {
        long long x, y;
        cin >> x >> y;
        v[i] = {x, y};
      }
    }
    reverse(v.begin(), v.end());
    for (long long i = 0; i < N; i++) par[i] = i;
    for (long long i = 0; i < q; i++) {
      if (v[i].second == -1) {
        long long x = v[i].first;
        ans.push_back(par[x]);
      } else {
        long long x = v[i].first;
        long long y = v[i].second;
        par[x] = par[y];
      }
    }
    reverse(ans.begin(), ans.end());
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  