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 = 500005;
  int getfa(int fa[], int x) { return fa[x] == x ? x : fa[x] = getfa(fa, fa[x]); }
  int q, n;
  int num[N], fa[N];
  int a[N];
  int main() {
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
      int op, x, y;
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d", &x);
        n++;
        if (num[x] == 0) {
          a[n] = x;
          num[x] = n;
          fa[n] = n;
        } else {
          fa[n] = num[x];
        }
      } else {
        scanf("%d%d", &x, &y);
        if (x != y && num[x] != 0) {
          if (num[y] == 0) {
            a[num[x]] = y;
            num[y] = num[x];
            num[x] = 0;
          } else {
            fa[num[x]] = num[y];
            a[num[x]] = y;
            num[x] = 0;
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", a[getfa(fa, i)]);
    }
    return 0;
  }
  