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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int inf = (int)0x3f3f3f3f3f3f3f3f, ninf = (int)0xc0c0c0c0c0c0c0c0;
  using ll = long long;
  using ld = long double;
  void solve() {
    const int N = (int)5e5 + 3;
    int q;
    cin >> q;
    vector<vector<int> > color(N);
    int id = 0;
    while (q--) {
      int t;
      cin >> t;
      if (t == 1) {
        int x;
        cin >> x;
        color[x].push_back(++id);
      } else {
        int x, y;
        cin >> x >> y;
        if (x == y) continue;
        if (((int)(color[y]).size()) < ((int)(color[x]).size()))
          swap(color[x], color[y]);
        color[y].insert(color[y].end(), color[x].begin(), color[x].end());
        color[x].clear();
      }
    }
    vector<int> ans(N);
    for (int i = 1; i <= N; ++i) {
      for (auto& x : color[i]) {
        ans[x] = i;
      }
    }
    for (int i = 1; i <= id; ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(0);
    int t = 1;
    while (t--) solve();
    return 0;
  }
  