#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  long long mx = 0;
  vector<long long> ty(t);
  vector<long long> a(t);
  vector<long long> b(t);
  for (long long i = 0; i < t; i++) {
    cin >> ty[i] >> a[i];
    mx = max(a[i], mx);
    if (ty[i] == 2) {
      cin >> b[i];
      mx = max(mx, b[i]);
    }
  }
  vector<long long> y(mx + 1);
  for (long long i = 0; i < mx + 1; i++) {
    y[i] = i;
  }
  vector<long long> ans;
  long long z = -1, k = -1;
  for (long long i = t - 1; i >= 0; i--) {
    if (ty[i] == 1) {
      ans.push_back(y[a[i]]);
    } else {
      y[a[i]] = y[b[i]];
    }
  }
  reverse(ans.begin(), ans.end());
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long mod = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  long long mx = 0;
  vector<long long> ty(t);
  vector<long long> a(t);
  vector<long long> b(t);
  for (long long i = 0; i < t; i++) {
    cin >> ty[i] >> a[i];
    mx = max(a[i], mx);
    if (ty[i] == 2) {
      cin >> b[i];
      mx = max(mx, b[i]);
    }
  }
  vector<long long> y(mx + 1);
  for (long long i = 0; i < mx + 1; i++) {
    y[i] = i;
  }
  vector<long long> ans;
  long long z = -1, k = -1;
  for (long long i = t - 1; i >= 0; i--) {
    if (ty[i] == 1) {
      ans.push_back(y[a[i]]);
    } else {
      y[a[i]] = y[b[i]];
    }
  }
  reverse(ans.begin(), ans.end());
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}

```