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 clang diagnostic push
  #pragma ide diagnostic ignored "cppcoreguidelines-narrowing-conversions"
  #pragma ide diagnostic ignored "hicpp-signed-bitwise"
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC optimize("no-stack-protector,fast-math")
  using namespace std;
  constexpr int N = 2e5 + 7;
  constexpr int M = 998244353;
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    long long q;
    cin >> q;
    vector<tuple<long long, long long, long long>> a(q);
    for (auto& [op, x, y] : a) {
      cin >> op >> x;
      if (op == 2) cin >> y;
    }
    vector<long long> ans;
    map<long long, long long> t;
    reverse(a.begin(), a.end());
    for (auto& [op, x, y] : a) {
      if (op == 1) {
        if (t.count(x)) {
          ans.emplace_back(t[x]);
        } else {
          ans.emplace_back(x);
        }
      } else {
        if (t.count(y)) {
          t[x] = t[y];
        } else {
          t[x] = y;
        }
      }
    }
    reverse(ans.begin(), ans.end());
    for (long long x : ans) cout << x << " ";
    cout << "\n";
    return 0;
  }
  #pragma clang diagnostic pop
  