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>
  #pragma gcc optimize("Ofast,no-stack-protector,tune=native")
  #pragma gcc optimize("sse,sse2,sse3,sse4,ssse3")
  #pragma gcc optimize("O2")
  #pragma gcc optimize("abm,mmx,avx,avx2,unroll-loops,fast-math,section-anchors")
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int MAXN = 5e5 + 5;
  vector<int> val[MAXN];
  void solve() {
    int q, r, x, y;
    cin >> q;
    int sz = 0;
    while (q--) {
      cin >> r;
      if (r == 1) {
        cin >> x;
        val[x].push_back(sz++);
      } else {
        cin >> x >> y;
        if (x == y) continue;
        if (val[x].size() > val[y].size()) swap(val[x], val[y]);
        for (auto c : val[x]) {
          val[y].push_back(c);
        }
        val[x].clear();
      }
    }
    vector<int> ans(sz);
    for (int i = 0; i < MAXN; i++) {
      for (auto c : val[i]) {
        ans[c] = i;
      }
    }
    for (auto c : ans) cout << c << ' ';
    cout << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
  }
  