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;
  void canhazfast() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  template <typename T>
  T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
  }
  template <typename T>
  T extgcd(T a, T b, T &x, T &y) {
    T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
    while (b) {
      T q = a / b;
      a %= b;
      swap(a, b);
      x0 -= q * x1;
      swap(x0, x1);
      y0 -= q * y1;
      swap(y0, y1);
    }
    x = x0;
    y = y0;
    return a;
  }
  int ctz(unsigned x) { return __builtin_ctz(x); }
  int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
  int clz(unsigned x) { return __builtin_clz(x); }
  int clzll(unsigned long long x) { return __builtin_clzll(x); }
  int popcnt(unsigned x) { return __builtin_popcount(x); }
  int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
  int bsr(unsigned x) { return 31 ^ clz(x); }
  int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
  int f[500016], x[500016], y[500016];
  int main() {
    canhazfast();
    int q;
    cin >> q;
    iota(f, f + 500016, 0);
    for (int i = 0; i < q; ++i) {
      int t;
      cin >> t >> x[i];
      if (t == 2) cin >> y[i];
    }
    for (int i = q - 1; i >= 0; --i) {
      if (y[i] == 0)
        x[i] = f[x[i]];
      else
        f[x[i]] = f[y[i]];
    }
    for (int i = 0; i < q; ++i) {
      if (y[i] == 0) cout << x[i] << ' ';
    }
    cout << '\n';
    return 0;
  }
  