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;
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  const long long N = 0;
  class dsu {
   public:
    vector<long long> parent;
    vector<long long> size;
    vector<long long> rank;
    vector<long long> val;
    explicit dsu(long long a) {
      parent.resize(a);
      size.resize(a);
      rank.resize(a);
      val.resize(a);
      for (long long i = 0; i < a; i++) {
        parent[i] = i;
        size[i] = 1;
        rank[i] = 0;
        val[i] = -1;
      }
    }
    long long par(long long i) {
      if (i == parent[i]) return i;
      return parent[i] = par(parent[i]);
    }
    bool unite(long long a, long long b, long long c) {
      a = par(a);
      b = par(b);
      if (a != b) {
        if (rank[a] > rank[b]) swap(a, b);
        parent[a] = b;
        size[b] += size[a];
        size[a] = 0;
        val[b] = c;
        val[a] = -1;
        if (rank[a] == rank[b]) rank[b]++;
        return true;
      }
      return false;
    }
  };
  void solve() {
    long long n;
    cin >> n;
    map<long long, long long> mp;
    dsu d(n + 5);
    long long cur = 0;
    long long temp = n;
    while (n--) {
      long long t;
      cin >> t;
      if (t == 1) {
        long long last;
        cin >> last;
        if (!mp.count(last)) {
          mp[last] = cur;
          d.val[cur] = last;
        } else {
          d.unite(cur, mp[last], last);
        }
        cur++;
      } else {
        long long x, y;
        cin >> x >> y;
        if (!mp.count(x)) continue;
        if (!mp.count(y)) {
          mp[y] = mp[x];
          long long par = d.par(mp[y]);
          d.val[par] = y;
        } else {
          long long xx = mp[x];
          long long yy = mp[y];
          d.unite(xx, yy, y);
        }
        if (x != y) mp.erase(x);
      }
    }
    for (long long i = 0; i < cur; i++) {
      long long par = d.par(i);
      cout << d.val[par] << " ";
    }
    cout << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    for (long long i = 1; i <= t; i++) {
      solve();
    }
  }
  