#include <bits/stdc++.h>
using namespace std;
void solve() {
  int64_t q;
  cin >> q;
  int64_t t, x, y;
  vector<vector<int64_t> > ans(q);
  for (int64_t i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      ans[i].push_back(x);
    } else {
      cin >> x >> y;
      ans[i].push_back(x);
      ans[i].push_back(y);
    }
  }
  ans.shrink_to_fit();
  map<int64_t, int64_t> m;
  stack<int64_t> s;
  for (int64_t i = q - 1; i >= 0; i--) {
    if (ans[i].size() == 2) {
      if (ans[i][0] != ans[i][1]) {
        if (m.count(ans[i][1])) {
          m[ans[i][0]] = m[ans[i][1]];
        } else {
          m[ans[i][0]] = ans[i][1];
        }
      }
    } else {
      int64_t p = ans[i][0];
      if (m.count(p)) {
        p = m[p];
      }
      s.push(p);
    }
  }
  while (!s.empty()) {
    cout << s.top() << " ";
    s.pop();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t test = 1;
  while (test--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
void solve() {
  int64_t q;
  cin >> q;
  int64_t t, x, y;
  vector<vector<int64_t> > ans(q);
  for (int64_t i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      ans[i].push_back(x);
    } else {
      cin >> x >> y;
      ans[i].push_back(x);
      ans[i].push_back(y);
    }
  }
  ans.shrink_to_fit();
  map<int64_t, int64_t> m;
  stack<int64_t> s;
  for (int64_t i = q - 1; i >= 0; i--) {
    if (ans[i].size() == 2) {
      if (ans[i][0] != ans[i][1]) {
        if (m.count(ans[i][1])) {
          m[ans[i][0]] = m[ans[i][1]];
        } else {
          m[ans[i][0]] = ans[i][1];
        }
      }
    } else {
      int64_t p = ans[i][0];
      if (m.count(p)) {
        p = m[p];
      }
      s.push(p);
    }
  }
  while (!s.empty()) {
    cout << s.top() << " ";
    s.pop();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t test = 1;
  while (test--) {
    solve();
  }
  return 0;
}

```