#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 5e5 + 100;
vector<int> pos[maxn];
int ans[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) {
    int n;
    cin >> n;
    int k = 0;
    for (int i = 1; i <= n; ++i) {
      int op;
      cin >> op;
      if (op == 1) {
        int x;
        cin >> x;
        pos[x].push_back(++k);
      } else {
        int x, y;
        cin >> x >> y;
        if (x == y) {
          continue;
        }
        if (pos[x].size() > pos[y].size()) {
          swap(pos[x], pos[y]);
        }
        for (auto item : pos[x]) {
          pos[y].push_back(item);
        }
        pos[x].clear();
      }
    }
    for (int i = 1; i < maxn; ++i) {
      for (auto item : pos[i]) {
        ans[item] = i;
      }
    }
    for (int i = 1; i <= k; ++i) {
      cout << ans[i] << (i == k ? "\n" : " ");
    }
  }
}

### 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;
const long long mod = 1e9 + 7;
const int maxn = 5e5 + 100;
vector<int> pos[maxn];
int ans[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) {
    int n;
    cin >> n;
    int k = 0;
    for (int i = 1; i <= n; ++i) {
      int op;
      cin >> op;
      if (op == 1) {
        int x;
        cin >> x;
        pos[x].push_back(++k);
      } else {
        int x, y;
        cin >> x >> y;
        if (x == y) {
          continue;
        }
        if (pos[x].size() > pos[y].size()) {
          swap(pos[x], pos[y]);
        }
        for (auto item : pos[x]) {
          pos[y].push_back(item);
        }
        pos[x].clear();
      }
    }
    for (int i = 1; i < maxn; ++i) {
      for (auto item : pos[i]) {
        ans[item] = i;
      }
    }
    for (int i = 1; i <= k; ++i) {
      cout << ans[i] << (i == k ? "\n" : " ");
    }
  }
}

```