#include <bits/stdc++.h>
using namespace std;
void showlist(list<int> g) {
  list<int>::iterator it;
  for (it = g.begin(); it != g.end(); it++) {
    cout << *it << " ";
  }
  cout << '\n';
}
void solve() {
  long long n;
  cin >> n;
  map<int, list<int> > ar;
  int k = 0;
  int t;
  int x;
  int y;
  while (n--) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      ar[x].push_back(k);
      k++;
    } else {
      cin >> x;
      cin >> y;
      if (x != y) {
        ar[y].splice(ar[y].begin(), ar[x]);
        ar[x] = {};
      }
    }
  }
  int aj[k];
  for (auto i : ar) {
    if (i.second.begin() != i.second.end()) {
      for (auto j : i.second) {
        aj[j] = i.first;
      }
    }
  }
  for (int i = 0; i < k; i++) {
    cout << aj[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void showlist(list<int> g) {
  list<int>::iterator it;
  for (it = g.begin(); it != g.end(); it++) {
    cout << *it << " ";
  }
  cout << '\n';
}
void solve() {
  long long n;
  cin >> n;
  map<int, list<int> > ar;
  int k = 0;
  int t;
  int x;
  int y;
  while (n--) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      ar[x].push_back(k);
      k++;
    } else {
      cin >> x;
      cin >> y;
      if (x != y) {
        ar[y].splice(ar[y].begin(), ar[x]);
        ar[x] = {};
      }
    }
  }
  int aj[k];
  for (auto i : ar) {
    if (i.second.begin() != i.second.end()) {
      for (auto j : i.second) {
        aj[j] = i.first;
      }
    }
  }
  for (int i = 0; i < k; i++) {
    cout << aj[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
  return 0;
}

```