#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
void solve() {
  map<int, vector<int> > pos;
  int arr[500005];
  int q, cnt1 = 0;
  cin >> q;
  while (q--) {
    int a, x, y;
    cin >> a;
    if (a == 1) {
      cin >> x;
      pos[x].push_back(cnt1);
      cnt1++;
    } else {
      cin >> x >> y;
      if (x != y) {
        if (pos[x].size() > pos[y].size()) {
          swap(pos[x], pos[y]);
        }
        for (int i = 0; i < pos[x].size(); i++) {
          pos[y].push_back(pos[x][i]);
        }
        pos[x].clear();
      }
    }
  }
  for (int i = 0; i < 500005; i++) {
    for (int j = 0; j < pos[i].size(); j++) {
      arr[pos[i][j]] = i;
    }
  }
  for (int i = 0; i < cnt1; i++) {
    cout << arr[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t, tc;
  t = 1, tc = 0;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create 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 mod = 1e9 + 7;
void solve() {
  map<int, vector<int> > pos;
  int arr[500005];
  int q, cnt1 = 0;
  cin >> q;
  while (q--) {
    int a, x, y;
    cin >> a;
    if (a == 1) {
      cin >> x;
      pos[x].push_back(cnt1);
      cnt1++;
    } else {
      cin >> x >> y;
      if (x != y) {
        if (pos[x].size() > pos[y].size()) {
          swap(pos[x], pos[y]);
        }
        for (int i = 0; i < pos[x].size(); i++) {
          pos[y].push_back(pos[x][i]);
        }
        pos[x].clear();
      }
    }
  }
  for (int i = 0; i < 500005; i++) {
    for (int j = 0; j < pos[i].size(); j++) {
      arr[pos[i][j]] = i;
    }
  }
  for (int i = 0; i < cnt1; i++) {
    cout << arr[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t, tc;
  t = 1, tc = 0;
  while (t--) {
    solve();
  }
}

```