#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int q, ans[N];
map<int, vector<int> > pos;
int main() {
  cin >> q;
  int idx = 0;
  for (int qq = 0; qq < q; qq++) {
    int t;
    cin >> t;
    int x, y;
    if (t == 1) {
      cin >> x;
      pos[x].push_back(idx);
      idx++;
    } else {
      cin >> x >> y;
      if (x == y) continue;
      if (pos[x].size() > pos[y].size()) pos[x].swap(pos[y]);
      for (int& i : pos[x]) pos[y].push_back(i);
      pos[x].clear();
    }
  }
  memset(ans, -1, sizeof ans);
  for (const auto& p : pos) {
    for (int i : p.second) ans[i] = p.first;
  }
  for (int i = 0; i < N; i++) {
    if (ans[i] != -1) cout << ans[i] << " ";
  }
  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;
const int N = 5e5 + 5;
int q, ans[N];
map<int, vector<int> > pos;
int main() {
  cin >> q;
  int idx = 0;
  for (int qq = 0; qq < q; qq++) {
    int t;
    cin >> t;
    int x, y;
    if (t == 1) {
      cin >> x;
      pos[x].push_back(idx);
      idx++;
    } else {
      cin >> x >> y;
      if (x == y) continue;
      if (pos[x].size() > pos[y].size()) pos[x].swap(pos[y]);
      for (int& i : pos[x]) pos[y].push_back(i);
      pos[x].clear();
    }
  }
  memset(ans, -1, sizeof ans);
  for (const auto& p : pos) {
    for (int i : p.second) ans[i] = p.first;
  }
  for (int i = 0; i < N; i++) {
    if (ans[i] != -1) cout << ans[i] << " ";
  }
  return 0;
}

```