#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int n, cnt, a[N], x[N], y[N], fa[N], ans[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < N; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      scanf("%d", &x[i]);
    else
      scanf("%d%d", &x[i], &y[i]);
  }
  for (int i = n; i > 0; i--) {
    if (a[i] == 1)
      ans[++cnt] = fa[x[i]];
    else
      fa[x[i]] = fa[y[i]];
  }
  for (int i = cnt; i > 0; i--) printf("%d ", ans[i]);
  printf("\n");
}

### Prompt

Develop a solution in cpp to the problem described below:

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 N = 5e5 + 5;
int n, cnt, a[N], x[N], y[N], fa[N], ans[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < N; i++) fa[i] = i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      scanf("%d", &x[i]);
    else
      scanf("%d%d", &x[i], &y[i]);
  }
  for (int i = n; i > 0; i--) {
    if (a[i] == 1)
      ans[++cnt] = fa[x[i]];
    else
      fa[x[i]] = fa[y[i]];
  }
  for (int i = cnt; i > 0; i--) printf("%d ", ans[i]);
  printf("\n");
}

```