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>
  const int N = 5e5;
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int my_rand(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
  int par[N + 5];
  void init() {
    for (long long int i = 1; i <= N; i++) par[i] = i;
  }
  int get(int n) {
    if (par[n] == n) return n;
    return par[n] = get(par[n]);
  }
  void uni(int a, int b) { par[a] = b; }
  int v[500005];
  int in[500005];
  void solve() {
    init();
    long long int nw = 1e5;
    long long int q;
    long long int i = 1;
    cin >> q;
    int x, y;
    while (q--) {
      long long int t;
      cin >> t;
      if (t == 1) {
        cin >> x;
        v[i] = x;
        if (in[x]) {
          uni(get(in[x]), i);
        } else {
          in[x] = i;
        }
        i++;
      } else {
        cin >> x >> y;
        int xin = in[x];
        if (xin == 0) continue;
        long long int yin = in[y];
        yin = get(yin);
        xin = get(xin);
        if (yin == 0) {
          yin = xin;
          v[yin] = y;
          in[y] = yin;
          in[x] = 0;
        } else {
          if (get(in[x]) == get(in[y])) continue;
          uni(get(in[x]), get(in[y]));
          in[x] = 0;
        }
      }
    }
    for (long long int j = 1; j <= i - 1; j++) {
      x = get(j);
      cout << v[x] << ' ';
    }
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
  }
  