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;
  template <class T>
  using vc = vector<T>;
  template <class T>
  using vvc = vc<vc<T>>;
  template <class T>
  void mkuni(vector<T>& v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
  }
  template <class T>
  void print(T x, int suc = 1) {
    cout << x;
    if (suc == 1)
      cout << '\n';
    else
      cout << ' ';
  }
  template <class T>
  void print(const vector<T>& v, int suc = 1) {
    for (int i = 0; i < v.size(); i++)
      print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
  }
  int dp[500050];
  pair<int, int> a[500050];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int q;
    cin >> q;
    for (int i = 1, op; i <= q; ++i) {
      cin >> op;
      if (op == 1) {
        int x;
        cin >> x;
        a[i] = {x, -1};
      } else {
        int x, y;
        cin >> x >> y;
        a[i] = {x, y};
      }
    }
    vector<int> ans;
    for (int i = q; i >= 1; --i) {
      if (a[i].second == -1) {
        if (dp[a[i].first] == 0)
          ans.push_back(a[i].first), dp[a[i].first] = a[i].first;
        else
          ans.push_back(dp[a[i].first]);
      } else {
        int x = a[i].first, y = a[i].second;
        if (dp[y] == 0) dp[y] = y;
        dp[x] = dp[y];
      }
    }
    reverse(ans.begin(), ans.end());
    print(ans);
  }
  