#include <bits/stdc++.h>
using namespace std;
const int mxn = 5e5 + 5;
int p[mxn], t[mxn], x[mxn], y[mxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  vector<int> a;
  for (int i = 1; i <= mxn - 5; i++) p[i] = i;
  for (int i = q; i >= 1; i--) {
    cin >> t[i] >> x[i];
    if (t[i] == 2) cin >> y[i];
  }
  for (int i = 1; i <= q; i++)
    if (t[i] == 1)
      a.push_back(p[x[i]]);
    else
      p[x[i]] = p[y[i]];
  for (int i = a.size() - 1; i >= 0; i--) cout << a[i] << " ";
}

### Prompt

Generate 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;
const int mxn = 5e5 + 5;
int p[mxn], t[mxn], x[mxn], y[mxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> q;
  vector<int> a;
  for (int i = 1; i <= mxn - 5; i++) p[i] = i;
  for (int i = q; i >= 1; i--) {
    cin >> t[i] >> x[i];
    if (t[i] == 2) cin >> y[i];
  }
  for (int i = 1; i <= q; i++)
    if (t[i] == 1)
      a.push_back(p[x[i]]);
    else
      p[x[i]] = p[y[i]];
  for (int i = a.size() - 1; i >= 0; i--) cout << a[i] << " ";
}

```