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 N = 1e6 + 5;
  struct Dat {
    int f, x, y;
    Dat() {}
    Dat(int f, int x, int y) {
      this->f = f;
      this->x = x;
      this->y = y;
    }
  };
  Dat dat[N];
  int fa[N];
  int find(int x);
  int ans[N];
  int dfn[N];
  int main() {
    int q;
    scanf("%d", &q);
    for (int i = 0; i < N; i++) {
      fa[i] = i;
      dfn[i] = N;
    }
    for (int i = 0; i < q; i++) {
      int f, x, y;
      scanf("%d%d", &f, &x);
      if (f == 2) {
        scanf("%d", &y);
      }
      dat[i] = Dat(f, x, y);
    }
    int cnt = 0;
    for (int i = q - 1; i >= 0; i--) {
      if (dat[i].f == 1) {
        ans[cnt++] = fa[dat[i].x];
      } else {
        fa[dat[i].x] = fa[dat[i].y];
      }
    }
    for (int i = cnt - 1; i >= 0; i--) {
      printf("%d%c", ans[i], i == 0 ? '\n' : ' ');
    }
    return 0;
  }
  int find(int x) {
    if (x == fa[x]) {
      return x;
    } else {
      if (dfn[fa[x]] > dfn[x]) {
        return fa[x] = find(fa[x]);
      }
      return fa[x];
    }
  }
  