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;
  void solve() {
    int64_t q;
    cin >> q;
    int64_t t, x, y;
    vector<vector<int64_t> > ans(q);
    for (int64_t i = 0; i < q; i++) {
      cin >> t;
      if (t == 1) {
        cin >> x;
        ans[i].push_back(x);
      } else {
        cin >> x >> y;
        ans[i].push_back(x);
        ans[i].push_back(y);
      }
    }
    ans.shrink_to_fit();
    map<int64_t, int64_t> m;
    stack<int64_t> s;
    for (int64_t i = q - 1; i >= 0; i--) {
      if (ans[i].size() == 2) {
        if (ans[i][0] != ans[i][1]) {
          if (m.count(ans[i][1])) {
            m[ans[i][0]] = m[ans[i][1]];
          } else {
            m[ans[i][0]] = ans[i][1];
          }
        }
      } else {
        int64_t p = ans[i][0];
        if (m.count(p)) {
          p = m[p];
        }
        s.push(p);
      }
    }
    while (!s.empty()) {
      cout << s.top() << " ";
      s.pop();
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int64_t test = 1;
    while (test--) {
      solve();
    }
    return 0;
  }
  