#include <bits/stdc++.h>
using namespace std;
int n, x, y;
int fa[500010];
vector<pair<int, pair<int, int>>> q;
stack<int> out;
void solve() {
  for (int i = 0; i < 500010; i++) fa[i] = i;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x == 1) {
      cin >> y;
      q.push_back({1, {y, 0}});
    } else {
      cin >> x >> y;
      q.push_back({2, {x, y}});
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (q[i].first == 1)
      out.push(fa[q[i].second.first]);
    else
      fa[q[i].second.first] = fa[q[i].second.second];
  }
  while (!out.empty()) {
    cout << out.top() << " ";
    out.pop();
  }
  cout << '\n';
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  solve();
  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;
int n, x, y;
int fa[500010];
vector<pair<int, pair<int, int>>> q;
stack<int> out;
void solve() {
  for (int i = 0; i < 500010; i++) fa[i] = i;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x == 1) {
      cin >> y;
      q.push_back({1, {y, 0}});
    } else {
      cin >> x >> y;
      q.push_back({2, {x, y}});
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (q[i].first == 1)
      out.push(fa[q[i].second.first]);
    else
      fa[q[i].second.first] = fa[q[i].second.second];
  }
  while (!out.empty()) {
    cout << out.top() << " ";
    out.pop();
  }
  cout << '\n';
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  solve();
  return 0;
}

```