#include <bits/stdc++.h>
using namespace std;
void print(vector<long long int> &arr) {
  long long int n = arr.size();
  for (long long int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << "\n";
}
void init_code() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void pre() {}
int main() {
  init_code();
  pre();
  long long int q;
  cin >> q;
  vector<vector<long long int>> queries;
  for (long long int i = 0; i < q; i++) {
    long long int type, x, y;
    cin >> type;
    if (type == 1) {
      cin >> x;
      vector<long long int> temp({type, x});
      queries.push_back(temp);
    } else {
      cin >> x >> y;
      vector<long long int> temp({type, x, y});
      queries.push_back(temp);
    }
  }
  vector<long long int> ans;
  unordered_map<long long int, long long int> mp;
  for (long long int i = q - 1; i >= 0; i--) {
    if (queries[i][0] == 1) {
      long long int x = queries[i][1];
      long long int y = x;
      if (mp.find(y) != mp.end()) {
        y = mp[y];
      }
      ans.push_back(y);
    } else {
      long long int x = queries[i][1];
      long long int y = queries[i][2];
      if (x != y) {
        mp[x] = mp.find(y) == mp.end() ? y : mp[y];
      }
    }
  }
  reverse(ans.begin(), ans.end());
  print(ans);
  return 0;
}

### 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;
void print(vector<long long int> &arr) {
  long long int n = arr.size();
  for (long long int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << "\n";
}
void init_code() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void pre() {}
int main() {
  init_code();
  pre();
  long long int q;
  cin >> q;
  vector<vector<long long int>> queries;
  for (long long int i = 0; i < q; i++) {
    long long int type, x, y;
    cin >> type;
    if (type == 1) {
      cin >> x;
      vector<long long int> temp({type, x});
      queries.push_back(temp);
    } else {
      cin >> x >> y;
      vector<long long int> temp({type, x, y});
      queries.push_back(temp);
    }
  }
  vector<long long int> ans;
  unordered_map<long long int, long long int> mp;
  for (long long int i = q - 1; i >= 0; i--) {
    if (queries[i][0] == 1) {
      long long int x = queries[i][1];
      long long int y = x;
      if (mp.find(y) != mp.end()) {
        y = mp[y];
      }
      ans.push_back(y);
    } else {
      long long int x = queries[i][1];
      long long int y = queries[i][2];
      if (x != y) {
        mp[x] = mp.find(y) == mp.end() ? y : mp[y];
      }
    }
  }
  reverse(ans.begin(), ans.end());
  print(ans);
  return 0;
}

```