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 MAXN = 1e6 + 10, LIM = 5e5;
  int q, op, x, y, tot, cur, fa[MAXN * 2], rt[MAXN], tag[MAXN * 2], col[MAXN * 2];
  int ans[MAXN];
  int Find(int x) {
    if (fa[x] == x) return x;
    return fa[x] = Find(fa[x]);
  }
  int main() {
    tot = LIM;
    for (int i = (1); i <= (tot); i++) fa[i] = i, rt[i] = i, col[i] = i;
    cin >> q;
    for (int i = (1); i <= (q); i++) {
      cin >> op;
      if (op == 1) {
        cin >> x;
        tot++;
        cur++;
        tag[tot] = cur;
        fa[tot] = Find(rt[x]);
      } else {
        cin >> x >> y;
        if (x == y) continue;
        fa[Find(rt[x])] = Find(rt[y]);
        tot++;
        fa[tot] = tot;
        rt[x] = tot;
        col[tot] = x;
      }
    }
    for (int i = (1); i <= (tot); i++) {
      if (!tag[i]) continue;
      int u = Find(i);
      ans[tag[i]] = col[u];
    }
    for (int i = (1); i <= (cur); i++) printf("%d ", ans[i]);
    return 0;
  }
  