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 main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long q;
    cin >> q;
    vector<long long> ret;
    vector<int> p(5e5 + 1, -1);
    function<int(int)> myFind = [&](int n) {
      if (p[n] == -1) {
        return n;
      }
      return p[n] = myFind(p[n]);
    };
    vector<tuple<long long, long long, long long> > query(q);
    for (int i = 0; i < q; ++i) {
      long long op, x, y;
      cin >> op;
      if (op == 1) {
        cin >> x;
      } else {
        cin >> x >> y;
      }
      query[i] = {op, x, y};
    }
    reverse((query).begin(), (query).end());
    map<long long, long long> m;
    for (auto [op, x, y] : query) {
      if (op == 1) {
        if (m.count(x)) {
          ret.push_back(m[x]);
        } else {
          ret.push_back(x);
        }
      } else {
        if (m.count(y)) {
          m[x] = m[y];
        } else {
          m[x] = y;
        }
      }
    }
    reverse((ret).begin(), (ret).end());
    for (auto e : ret) {
      cout << e << ' ';
    }
    cout << '\n';
  }
  