#include <bits/stdc++.h>
using namespace std;
const int N = 500005;
int getfa(int fa[], int x) { return fa[x] == x ? x : fa[x] = getfa(fa, fa[x]); }
int q, n;
int num[N], fa[N];
int a[N];
int main() {
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int op, x, y;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &x);
      n++;
      if (num[x] == 0) {
        a[n] = x;
        num[x] = n;
        fa[n] = n;
      } else {
        fa[n] = num[x];
      }
    } else {
      scanf("%d%d", &x, &y);
      if (x != y && num[x] != 0) {
        if (num[y] == 0) {
          a[num[x]] = y;
          num[y] = num[x];
          num[x] = 0;
        } else {
          fa[num[x]] = num[y];
          a[num[x]] = y;
          num[x] = 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[getfa(fa, i)]);
  }
  return 0;
}

### 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 int N = 500005;
int getfa(int fa[], int x) { return fa[x] == x ? x : fa[x] = getfa(fa, fa[x]); }
int q, n;
int num[N], fa[N];
int a[N];
int main() {
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int op, x, y;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &x);
      n++;
      if (num[x] == 0) {
        a[n] = x;
        num[x] = n;
        fa[n] = n;
      } else {
        fa[n] = num[x];
      }
    } else {
      scanf("%d%d", &x, &y);
      if (x != y && num[x] != 0) {
        if (num[y] == 0) {
          a[num[x]] = y;
          num[y] = num[x];
          num[x] = 0;
        } else {
          fa[num[x]] = num[y];
          a[num[x]] = y;
          num[x] = 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[getfa(fa, i)]);
  }
  return 0;
}

```