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 int M = 500005;
  long long int N = 1e9 + 7;
  long long int Z = 998244353;
  long long int power(long long int x, unsigned int y, int p) {
    long long int res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long int modinv(long long int x, int p) { return (power(x, p - 2, p)); }
  bool compare(const pair<int, int> &p1, const pair<int, int> &p2) {
    return (p1.first < p2.first);
  }
  void precompute() {}
  int parent[M];
  map<int, int> mp, rmp;
  int find(int x) {
    if (x == parent[x]) {
      return x;
    }
    return parent[x] = find(parent[x]);
  }
  void merge(int x, int y) {
    if (x == y) {
      return;
    }
    if (mp.find(x) == mp.end()) {
      return;
    }
    int px = find(mp[x]);
    if (mp.find(y) == mp.end()) {
      rmp[px] = y;
      mp.erase(x);
      mp[y] = px;
      return;
    }
    int py = find(mp[y]);
    mp.erase(x);
    parent[px] = py;
  }
  void solve() {
    int q;
    cin >> q;
    int sz = 0;
    while (q--) {
      int type, x, y;
      cin >> type;
      if (type == 1) {
        cin >> x;
        if (mp.find(x) != mp.end()) {
          parent[sz] = mp[x];
        } else {
          mp[x] = sz;
          rmp[sz] = x;
          parent[sz] = sz;
        }
        sz++;
      } else {
        cin >> x >> y;
        merge(x, y);
      }
    }
    for (int i = 0; i < sz; i++) {
      cout << rmp[find(i)] << " ";
    }
    cout << "\n";
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    int tt = t;
    precompute();
    while (t--) {
      solve();
    }
    return 0;
  }
  