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("Ofast")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  struct dsu {
    int head[500005], size[500005], value[500005];
    dsu() {
      memset(head, -1, sizeof(head));
      for (int i = 0; i <= 500000; ++i) value[i] = i;
      for (int i = 0; i <= 500000; ++i) size[i] = 1;
    }
    int find_head(int node) {
      while (head[node] != -1) node = head[node];
      return node;
    }
    int merge(int a, int b) {
      int x = find_head(a), y = find_head(b);
      if (x != y) {
        if (size[x] > size[y])
          value[x] = value[y], size[x] += size[y], head[y] = x;
        else
          size[y] += size[x], head[x] = y;
      }
      return x != y;
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int q;
    cin >> q;
    pair<bool, pair<int, int>> queries[q];
    int arr_size = 0;
    for (int i = 0; i < q; ++i) {
      int t, x, y;
      cin >> t;
      if (t == 1)
        cin >> x, queries[i] = make_pair(0, make_pair(x, 0)), ++arr_size;
      else
        cin >> x >> y, queries[i] = make_pair(1, make_pair(x, y));
    }
    reverse(queries, queries + q);
    int res[arr_size], idx = 0;
    int value[500005];
    memset(value, -1, sizeof(value));
    for (int i = 0; i < q; ++i) {
      if (queries[i].first) {
        value[queries[i].second.first] = value[queries[i].second.second];
        if (value[queries[i].second.first] == -1)
          value[queries[i].second.first] = queries[i].second.second;
      } else {
        res[idx++] = value[queries[i].second.first];
        if (res[idx - 1] == -1) res[idx - 1] = queries[i].second.first;
      }
    }
    reverse(res, res + arr_size);
    for (int i = 0; i < arr_size; ++i) cout << res[i] << " ";
    cout << "\n";
    return 0;
  }
  