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;
  using ll = long long;
  using db = long double;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <typename T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  string to_string(char c) { return string(1, c); }
  string to_string(const char* s) { return (string)s; }
  string to_string(const string& s) { return '"' + s + '"'; }
  string to_string(bool b) { return (b ? "true" : "false"); }
  string to_string(vector<bool> v) {
    string res = "{";
    for (ll i = (0); i < (int((v).size())); ++i) res += char('0' + v[i]);
    res += "}";
    return res;
  }
  template <size_t SZ>
  string to_string(bitset<SZ> b) {
    string res;
    for (ll i = (0); i < (SZ); ++i) res += char('0' + b[i]);
    return res;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p);
  template <typename A>
  string to_string(A v) {
    bool first = 1;
    string res = "{";
    for (const auto& x : v) {
      if (!first) res += ", ";
      first = 0;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  void dbgOut() { cerr << "]" << endl; }
  template <typename H, typename... T>
  void dbgOut(H h, T... t) {
    cerr << to_string(h);
    if (sizeof...(t)) cerr << ", ";
    dbgOut(t...);
  }
  void solve() {
    int q;
    cin >> q;
    vector<pair<int, int> > queries;
    while (q--) {
      int k;
      cin >> k;
      if (k == 1) {
        int x;
        cin >> x;
        queries.push_back({x, -1});
      } else {
        int x, y;
        cin >> x >> y;
        queries.push_back({x, y});
      }
    }
    reverse((queries).begin(), (queries).end());
    0;
    vector<int> ans;
    map<int, int> final;
    for (auto x : queries) {
      if (x.second == -1) {
        if (final.find(x.first) == final.end())
          ans.push_back(x.first);
        else
          ans.push_back(final[x.first]);
      } else {
        if (final.find(x.second) == final.end())
          final[x.first] = x.second;
        else
          final[x.first] = final[x.second];
      }
    }
    reverse((ans).begin(), (ans).end());
    for (auto x : ans) {
      cout << x << " ";
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
  }
  