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;
  constexpr long long MAXX = 2e18;
  constexpr long double EPS = 1e-9;
  constexpr long long P = 1000000007;
  long long mod_add(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mod_mul(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long mod_sub(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  template <typename T>
  ostream& operator<<(ostream& output, const vector<T>& data) {
    for (const T& x : data) output << x << " ";
    return output;
  }
  template <typename T>
  istream& operator>>(istream& input, vector<T>& data) {
    for (auto& item : data) {
      input >> item;
    }
    return input;
  }
  vector<bool> seive(long long n) {
    vector<bool> arr(n + 1, true);
    for (int i = 2; i * i <= n; i++) {
      if (arr[i]) {
        for (int j = i * i; j <= n; j += i) {
          arr[j] = false;
        }
      }
    }
    return arr;
  }
  vector<long long> lp;
  vector<long long> seive_op(long long n) {
    vector<long long> prime;
    lp.assign(n + 1, 0);
    long long m = 0;
    for (int i = 2; i <= n; i++) {
      if (lp[i] == 0) {
        lp[i] = i;
        prime.push_back(i);
        m++;
      }
      for (int j = 0; (j < m) && (prime[j] <= lp[i]) && (i * prime[j] <= n);
           j++) {
        lp[i * prime[j]] = prime[j];
      }
    }
    return prime;
  }
  map<long long, long long> primefactorisation(long long n) {
    map<long long, long long> mp;
    while (n != 1) {
      mp[lp[n]]++;
      n /= (lp[n]);
    }
    return mp;
  }
  long long mod_exp(long long x, long long y, long long p) {
    long long ans = 1LL, r = 1LL;
    x %= p;
    while (r > 0 && r <= y) {
      if (r & y) {
        ans *= x;
        ans %= p;
      }
      r = (r << 1LL);
      x *= x;
      x %= p;
    }
    return ans;
  }
  long long mod_inv(long long n, long long p) { return mod_exp(n, p - 2LL, p); }
  bool isprime(long long x) {
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        return 0;
      }
    }
    return 1;
  }
  bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
    return (p1.first > p2.first);
  }
  vector<long long> fact;
  void pre_fermat() {
    fact.assign(1000001, 1);
    fact[0] = 1;
    for (int i = 1; i <= (1000000); i++) {
      fact[i] = mod_mul(fact[i - 1], i, P);
    }
  }
  long long ncr(long long n, long long r, long long p) {
    if (n < r) {
      return 0LL;
    }
    if (r == 0) {
      return 1LL;
    }
    return (fact[n] * mod_inv(fact[r], p) % p * mod_inv(fact[n - r], p) % p) % p;
  }
  vector<long long> bfs(vector<vector<long long>>& edge, long long start,
                        long long n) {
    long long curr = start;
    vector<bool> visited(n + 1, 0);
    vector<long long> lvl(n + 1);
    lvl[curr] = 0;
    queue<long long> qq;
    qq.push(curr);
    visited[curr] = 1;
    while (!qq.empty()) {
      curr = qq.front();
      for (long long x : edge[curr]) {
        if (!visited[x]) {
          lvl[x] = 1 + lvl[curr];
          qq.push(x);
          visited[x] = 1;
        }
      }
      qq.pop();
    }
    return lvl;
  }
  void dfs(vector<vector<long long>>& edge, long long curr, long long prev) {
    for (long long x : edge[curr]) {
      if (x != prev) {
        dfs(edge, x, curr);
      }
    }
  }
  vector<string> bor = {"YES", "NO"};
  class stk {
   public:
    long long parent;
    long long rank;
  };
  class Dsu {
   public:
    unordered_map<long long, stk> parent;
    void init(long long x) {
      stk temp;
      temp.parent = x;
      temp.rank = 0;
      parent[x] = temp;
    }
    long long find(long long x) {
      if (parent[x].parent == x) {
        return x;
      }
      return (parent[x].parent = find(parent[x].parent));
    }
    void Union(long long x, long long y) {
      long long p1 = find(x);
      long long p2 = find(y);
      if (p1 == p2) {
        return;
      }
      if (parent[p1].rank >= parent[p2].rank) {
        parent[p2].parent = p1;
        parent[p1].rank++;
      } else {
        parent[p1].parent = p2;
        parent[p2].rank++;
      }
    }
  };
  void solve() {
    vector<long long> ans;
    long long q;
    cin >> q;
    vector<pair<long long, pair<long long, long long>>> arr(q);
    for (int i = 0; i < q; i++) {
      long long a;
      cin >> a;
      arr[i].first = a;
      if (a == 1) {
        long long b;
        cin >> b;
        arr[i].second.first = b;
      } else {
        long long x, y;
        cin >> x >> y;
        arr[i].second = {x, y};
      }
    }
    vector<long long> paint(500001);
    for (int i = 1; i <= (500000); i++) paint[i] = i;
    for (int i = q - 1; i >= 0; i--) {
      if (arr[i].first == 1) {
        long long temp = arr[i].second.first;
        ans.push_back(paint[temp]);
      } else {
        long long x = arr[i].second.first;
        long long y = arr[i].second.second;
        paint[x] = paint[y];
      }
    }
    reverse(ans.begin(), ans.end());
    cout << ans << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  