#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5;
int mp[maxn];
vector<int> pos[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < maxn; i++) mp[i] = i;
  int n = 0;
  int Q;
  cin >> Q;
  while (Q--) {
    int t, x, y;
    cin >> t >> x;
    if (t == 1) {
      pos[mp[x]].push_back(++n);
      continue;
    }
    cin >> y;
    if (x == y) continue;
    if (pos[mp[x]].size() > pos[mp[y]].size()) {
      swap(mp[x], mp[y]);
    }
    for (int i : pos[mp[x]]) pos[mp[y]].push_back(i);
    pos[mp[x]].clear();
    pos[mp[x]].shrink_to_fit();
  }
  vector<int> ans(n);
  for (int i = 1; i < maxn; i++)
    for (int j : pos[mp[i]]) ans[j - 1] = i;
  for (int i : ans) cout << i << ' ';
  cout << endl;
  return 0;
}

### Prompt

Generate 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;
const int maxn = 5e5 + 5;
int mp[maxn];
vector<int> pos[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < maxn; i++) mp[i] = i;
  int n = 0;
  int Q;
  cin >> Q;
  while (Q--) {
    int t, x, y;
    cin >> t >> x;
    if (t == 1) {
      pos[mp[x]].push_back(++n);
      continue;
    }
    cin >> y;
    if (x == y) continue;
    if (pos[mp[x]].size() > pos[mp[y]].size()) {
      swap(mp[x], mp[y]);
    }
    for (int i : pos[mp[x]]) pos[mp[y]].push_back(i);
    pos[mp[x]].clear();
    pos[mp[x]].shrink_to_fit();
  }
  vector<int> ans(n);
  for (int i = 1; i < maxn; i++)
    for (int j : pos[mp[i]]) ans[j - 1] = i;
  for (int i : ans) cout << i << ' ';
  cout << endl;
  return 0;
}

```