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);
    map<long long, long long> m;
    for (int i = 1; i <= 5 * pow(10, 5); i++) {
      m[i] = i;
    }
    long long q;
    cin >> q;
    vector<pair<long long, pair<long long, long long>>> v;
    long long n = 0;
    for (int i = 0; i < q; i++) {
      long long inp1;
      cin >> inp1;
      if (inp1 == 1) {
        n++;
        long long inp2, inp3;
        cin >> inp2;
        inp3 = inp2;
        v.push_back({inp1, {inp2, inp3}});
      } else {
        long long inp2, inp3;
        cin >> inp2 >> inp3;
        v.push_back({inp1, {inp2, inp3}});
      }
    }
    long long ans[n];
    long long m1 = n;
    n--;
    set<long long> st;
    for (int i = q - 1; i >= 0; i--) {
      if (v[i].first == 1) {
        ans[n] = m[v[i].second.first];
        n--;
      } else {
        if (st.find(v[i].second.second) != st.end()) {
          m[v[i].second.first] = m[v[i].second.second];
          st.insert(v[i].second.first);
          continue;
        } else {
          m[v[i].second.first] = m[v[i].second.second];
          st.insert(v[i].second.first);
        }
      }
    }
    for (int i = 0; i < m1; i++) {
      cout << ans[i] << " ";
    }
    return 0;
  }
  