#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int par[N];
int main() {
  int n;
  cin >> n;
  stack<vector<int>> s;
  for (int i = 0; i < n; i++) {
    int t, x, y;
    cin >> t;
    vector<int> v;
    v.push_back(t);
    if (t == 1) {
      cin >> x;
      v.push_back(x);
    } else {
      cin >> x >> y;
      v.push_back(x);
      v.push_back(y);
    }
    s.push(v);
  }
  iota(par, par + N, 0);
  vector<int> ans;
  while (!s.empty()) {
    vector<int> v = s.top();
    s.pop();
    int t = v[0];
    if (t == 1) {
      int x = v[1];
      ans.push_back(par[x]);
    } else {
      int x = v[1], y = v[2];
      par[x] = par[y];
    }
  }
  reverse(ans.begin(), ans.end());
  for (int i : ans) cout << i << " ";
  cout << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 par[N];
int main() {
  int n;
  cin >> n;
  stack<vector<int>> s;
  for (int i = 0; i < n; i++) {
    int t, x, y;
    cin >> t;
    vector<int> v;
    v.push_back(t);
    if (t == 1) {
      cin >> x;
      v.push_back(x);
    } else {
      cin >> x >> y;
      v.push_back(x);
      v.push_back(y);
    }
    s.push(v);
  }
  iota(par, par + N, 0);
  vector<int> ans;
  while (!s.empty()) {
    vector<int> v = s.top();
    s.pop();
    int t = v[0];
    if (t == 1) {
      int x = v[1];
      ans.push_back(par[x]);
    } else {
      int x = v[1], y = v[2];
      par[x] = par[y];
    }
  }
  reverse(ans.begin(), ans.end());
  for (int i : ans) cout << i << " ";
  cout << '\n';
}

```