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 to[500000], of[500000], _index[500000];
  vector<int> v;
  int cnt;
  int find(int i) {
    if (_index[i] == i) return i;
    return _index[i] = find(_index[i]);
  }
  void add(int x) {
    if (to[find(of[x])] != x) {
      cnt++;
      to[cnt] = x;
      of[x] = cnt;
      v.push_back(cnt);
    } else {
      v.push_back(of[x]);
    }
  }
  void change(int x, int y) {
    if (x == y) return;
    if (to[find(of[y])] != y) {
      cnt++;
      of[y] = cnt;
      to[cnt] = y;
    }
    int pa = find(of[x]), pb = find(of[y]);
    if (to[pa] != x) return;
    _index[pa] = pb;
  }
  void print() {
    for (auto x : v) cout << to[find(x)] << " ";
  }
  void solve() {
    int q;
    cin >> q;
    for (int i = 1; i <= q; i++) _index[i] = i;
    int t, x, y;
    cnt = 0;
    while (q--) {
      cin >> t;
      if (t == 1) {
        cin >> x;
        add(x);
      } else {
        cin >> x >> y;
        change(x, y);
      }
    }
    print();
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) {
      solve();
      cout << "\n";
    }
    return 0;
  }
  