#include <bits/stdc++.h>
using namespace std;
const long long N = 5e5 + 10;
long long q;
long long st[N], ed[N], last[N], ans[N], cnt;
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> q;
  while (q--) {
    long long op;
    long long x;
    cin >> op >> x;
    if (op == 1) {
      if (!st[x]) {
        st[x] = ed[x] = ++cnt;
      } else {
        last[++cnt] = ed[x];
        ed[x] = cnt;
      }
    } else {
      long long y;
      cin >> y;
      if (x == y || !st[x]) continue;
      if (!st[y]) {
        st[y] = st[x];
        ed[y] = ed[x];
      } else {
        last[st[x]] = ed[y];
        ed[y] = ed[x];
      }
      st[x] = ed[x] = 0;
    }
  }
  for (long long i = (1); i <= (5e5); i++) {
    for (long long j = ed[i]; j; j = last[j]) ans[j] = i;
  }
  for (long long i = (1); i <= (cnt); i++) cout << ans[i] << " ";
}

### Prompt

Please provide a Cpp coded solution 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 long long N = 5e5 + 10;
long long q;
long long st[N], ed[N], last[N], ans[N], cnt;
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> q;
  while (q--) {
    long long op;
    long long x;
    cin >> op >> x;
    if (op == 1) {
      if (!st[x]) {
        st[x] = ed[x] = ++cnt;
      } else {
        last[++cnt] = ed[x];
        ed[x] = cnt;
      }
    } else {
      long long y;
      cin >> y;
      if (x == y || !st[x]) continue;
      if (!st[y]) {
        st[y] = st[x];
        ed[y] = ed[x];
      } else {
        last[st[x]] = ed[y];
        ed[y] = ed[x];
      }
      st[x] = ed[x] = 0;
    }
  }
  for (long long i = (1); i <= (5e5); i++) {
    for (long long j = ed[i]; j; j = last[j]) ans[j] = i;
  }
  for (long long i = (1); i <= (cnt); i++) cout << ans[i] << " ";
}

```