#include <bits/stdc++.h>
using namespace std;
using P = pair<int, int>;
int main() {
  int q;
  cin >> q;
  vector<P> v(q);
  for (int qi = 0; qi < q; qi++) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      v[qi] = {x, 0};
    } else {
      int x, y;
      cin >> x >> y;
      v[qi] = {-x, y};
    }
  }
  reverse(v.begin(), v.end());
  map<int, int> to;
  vector<int> ans;
  for (auto now : v) {
    if (now.first >= 0) {
      if (to.count(now.first)) {
        now.first = to[now.first];
      }
      ans.push_back(now.first);
    } else {
      now.first *= -1;
      if (to.count(now.second)) {
        to[now.first] = to[now.second];
      } else {
        to[now.first] = now.second;
      }
    }
  }
  reverse(ans.begin(), ans.end());
  cout << ans[0];
  for (int i = 1; i < (int)ans.size(); i++) {
    cout << " " << ans[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
using P = pair<int, int>;
int main() {
  int q;
  cin >> q;
  vector<P> v(q);
  for (int qi = 0; qi < q; qi++) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      v[qi] = {x, 0};
    } else {
      int x, y;
      cin >> x >> y;
      v[qi] = {-x, y};
    }
  }
  reverse(v.begin(), v.end());
  map<int, int> to;
  vector<int> ans;
  for (auto now : v) {
    if (now.first >= 0) {
      if (to.count(now.first)) {
        now.first = to[now.first];
      }
      ans.push_back(now.first);
    } else {
      now.first *= -1;
      if (to.count(now.second)) {
        to[now.first] = to[now.second];
      } else {
        to[now.first] = now.second;
      }
    }
  }
  reverse(ans.begin(), ans.end());
  cout << ans[0];
  for (int i = 1; i < (int)ans.size(); i++) {
    cout << " " << ans[i];
  }
  cout << endl;
  return 0;
}

```