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("O3")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  long long inv[601010];
  long long p[601010];
  long long num[601010];
  long long arr[601010];
  long long find(long long x) {
    if (p[x] == x) return x;
    return p[x] = find(p[x]);
  }
  long long uni(long long a, long long b) {
    a = find(a);
    b = find(b);
    if (a == b) return a;
    if (num[a] > num[b]) swap(a, b);
    num[b] = num[a] + num[b];
    p[a] = b;
    return b;
  }
  void solve() {
    long long Q;
    cin >> Q;
    long long i;
    long long cnt = 0;
    long long t, a, b;
    for (i = 1; i <= Q; i++) {
      cin >> t >> a;
      if (t == 2) cin >> b;
      if (t == 1) {
        cnt++;
        p[cnt] = cnt;
        num[cnt] = 1;
        if (inv[a]) {
          p[cnt] = inv[a];
          num[inv[a]]++;
        } else {
          arr[cnt] = a;
          inv[a] = cnt;
        }
      } else {
        if (a == b) continue;
        if (inv[a] == 0) continue;
        if (inv[b]) {
          long long x = inv[a];
          long long y = inv[b];
          long long u = uni(x, y);
          inv[a] = 0;
          inv[b] = u;
          arr[u] = b;
        } else {
          long long x = inv[a];
          arr[x] = b;
          inv[a] = 0;
          inv[b] = x;
        }
      }
    }
    for (i = 1; i <= cnt; i++) {
      long long x = find(i);
      cout << arr[x] << ' ';
    }
    cout << '\n';
  }
  void init() {}
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    long long T;
    T = 1;
    while (T--) {
      init();
      solve();
    }
  }
  