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() {
    long long int n, i, j, k;
    cin >> n;
    map<long long int, long long int> m;
    long long int l = 5 * 100001;
    for (i = 1; i < l; i++) {
      m[i] = i;
    }
    vector<long long int> x(l);
    vector<pair<long long int, long long int>> y(l);
    vector<long long int> ans;
    long long int t = n;
    long long int a, b, c;
    for (i = 0; i < t; i++) {
      cin >> a >> b;
      if (a == 2) {
        cin >> c;
        x[i] = 0;
        y[i] = make_pair(b, c);
      } else {
        x[i] = b;
        y[i] = make_pair(0, 0);
      }
    }
    t = n;
    for (i = t - 1; i >= 0; i--) {
      if (x[i] != 0) {
        ans.push_back(m[x[i]]);
      } else {
        pair<long long int, long long int> p = y[i];
        m[p.first] = m[p.second];
      }
    }
    reverse(ans.begin(), ans.end());
    for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
  }
  int main() {
    long long int t = 1, i;
    while (t--) {
      solve();
    }
    return 0;
  }
  