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>
  #pragma GCC optimize(3)
  using namespace std;
  inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  inline int qpow(int a, int n) {
    int ans = 1;
    while (n) {
      if (n & 1) {
        ans *= a;
      }
      a *= a;
      n >>= 1;
    }
    return ans;
  }
  inline long long _qpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  inline long long ksm(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  const int maxnum = 5e5 + 5;
  struct que {
    int op, x, y;
  };
  vector<int> ans;
  vector<que> v;
  int f[maxnum];
  bool vis[maxnum];
  int op, x, y, q;
  void slove() {
    cin >> q;
    while (q--) {
      cin >> op;
      if (op == 1) {
        cin >> x;
        v.push_back({1, x, 0});
      } else {
        cin >> x >> y;
        v.push_back({2, x, y});
      }
    }
    reverse(v.begin(), v.end());
    for (int i = 0; i < maxnum; i++) f[i] = i;
    for (auto q : v) {
      op = q.op, x = q.x, y = q.y;
      if (op == 1)
        ans.push_back(f[x]);
      else
        f[x] = f[y];
    }
    reverse(ans.begin(), ans.end());
    for (int x : ans) cout << x << " ";
    cout << "\n";
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    ;
    slove();
    return 0;
  }
  