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;
  void solve() {
    long long int n = ({
      long long int tito;
      cin >> tito;
      tito;
    });
    vector<vector<long long int>> v;
    map<long long int, vector<long long int>> mm;
    long long int pos = 0;
    for (long long int i = 0; i < n; i++) {
      long long int a = ({
        long long int tito;
        cin >> tito;
        tito;
      });
      long long int b, c;
      if (a == 1) {
        b = ({
          long long int tito;
          cin >> tito;
          tito;
        });
        c = 0;
      } else {
        b = ({
          long long int tito;
          cin >> tito;
          tito;
        });
        c = ({
          long long int tito;
          cin >> tito;
          tito;
        });
      }
      vector<long long int> vv;
      vv.push_back(a);
      vv.push_back(b);
      vv.push_back(c);
      v.push_back(vv);
    }
    vector<long long int> ans;
    vector<long long int> aa;
    for (long long int i = 0; i < 5 * 1e5 + 5; i++) {
      aa.push_back(i);
    }
    for (long long int i = 0; i < n; i++) {
      vector<long long int> vv = v[n - 1 - i];
      long long int a = vv[0];
      long long int b = vv[1];
      long long int c = vv[2];
      if (a == 1) {
        ans.push_back(aa[b]);
      } else {
        aa[b] = aa[c];
      }
    }
    long long int nn = ans.size();
    for (long long int i = 0; i < nn; i++) {
      cout << ans[nn - 1 - i] << " ";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t = 1;
    while (t--) solve();
    return 0;
  }
  