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;
  const long long int mma = 1e6 + 10;
  const long long int mod = 1e9 + 7;
  long long int par[mma];
  vector<long long int> g[mma];
  long long int get_par(long long int x) {
    if (x == par[x]) {
      return x;
    }
    return par[x] = get_par(par[x]);
  }
  void solve() {
    long long int q;
    cin >> q;
    for (int i = 0; i < mma; i++) {
      par[i] = i;
    }
    long long int k = 0;
    for (int i = 0; i < q; i++) {
      long long int u;
      cin >> u;
      if (u == 1) {
        long long int x;
        cin >> x;
        g[i].push_back(x);
        k++;
      } else {
        long long int x, y;
        cin >> x >> y;
        g[i].push_back(x);
        g[i].push_back(y);
      }
    }
    long long int a[k];
    long long int u = 0;
    for (int i = q - 1; i >= 0; i--) {
      if (g[i].size() == 2) {
        par[g[i][0]] = par[g[i][1]];
      } else {
        a[k - u - 1] = par[g[i][0]];
        u++;
      }
    }
    for (int i = 0; i < k; i++) {
      cout << a[i] << " ";
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
  }
  