#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10, LIM = 5e5;
int q, op, x, y, tot, cur, fa[MAXN * 2], rt[MAXN], tag[MAXN * 2], col[MAXN * 2];
int ans[MAXN];
int Find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = Find(fa[x]);
}
int main() {
  tot = LIM;
  for (int i = (1); i <= (tot); i++) fa[i] = i, rt[i] = i, col[i] = i;
  cin >> q;
  for (int i = (1); i <= (q); i++) {
    cin >> op;
    if (op == 1) {
      cin >> x;
      tot++;
      cur++;
      tag[tot] = cur;
      fa[tot] = Find(rt[x]);
    } else {
      cin >> x >> y;
      if (x == y) continue;
      fa[Find(rt[x])] = Find(rt[y]);
      tot++;
      fa[tot] = tot;
      rt[x] = tot;
      col[tot] = x;
    }
  }
  for (int i = (1); i <= (tot); i++) {
    if (!tag[i]) continue;
    int u = Find(i);
    ans[tag[i]] = col[u];
  }
  for (int i = (1); i <= (cur); i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Generate 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 MAXN = 1e6 + 10, LIM = 5e5;
int q, op, x, y, tot, cur, fa[MAXN * 2], rt[MAXN], tag[MAXN * 2], col[MAXN * 2];
int ans[MAXN];
int Find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = Find(fa[x]);
}
int main() {
  tot = LIM;
  for (int i = (1); i <= (tot); i++) fa[i] = i, rt[i] = i, col[i] = i;
  cin >> q;
  for (int i = (1); i <= (q); i++) {
    cin >> op;
    if (op == 1) {
      cin >> x;
      tot++;
      cur++;
      tag[tot] = cur;
      fa[tot] = Find(rt[x]);
    } else {
      cin >> x >> y;
      if (x == y) continue;
      fa[Find(rt[x])] = Find(rt[y]);
      tot++;
      fa[tot] = tot;
      rt[x] = tot;
      col[tot] = x;
    }
  }
  for (int i = (1); i <= (tot); i++) {
    if (!tag[i]) continue;
    int u = Find(i);
    ans[tag[i]] = col[u];
  }
  for (int i = (1); i <= (cur); i++) printf("%d ", ans[i]);
  return 0;
}

```