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 INF = 1E16;
  const long double EPS = 1E-9;
  const long double PI = 3.1415926535897932384626433832795;
  const int SIZE = 10000000;
  long double _abs(long double a) {
    if (a < 0) return -a;
    return a;
  }
  long long _abss(long long a) {
    if (a < 0) return -a;
    return a;
  }
  long long gcd(long long a, long long b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  void solve() {
    long long k;
    cin >> k;
    vector<long long> a(k), x(k), y(k);
    for (long long i = 0; i < long long(k); ++i) {
      cin >> a[i];
      if (a[i] == 1) {
        cin >> x[i];
      } else {
        cin >> x[i] >> y[i];
      }
    }
    vector<long long> ans, p(SIZE, 0);
    for (long long i = 0; i < long long(SIZE); ++i) {
      p[i] = i;
    }
    for (long long i = long long(k) - 1; i >= 0; --i) {
      if (a[i] == 1) {
        ans.push_back(p[x[i]]);
      } else {
        p[x[i]] = p[y[i]];
      }
    }
    reverse((ans).begin(), (ans).end());
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << endl;
    return;
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  