#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5, INF = 0x3f3f3f3f;
int h[maxn], t[maxn], e[maxn], ne[maxn], cur, idx;
int ans[maxn];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  memset(h, -1, sizeof h);
  memset(ne, -1, sizeof ne);
  int n;
  cin >> n;
  while (n--) {
    int op, x, y;
    cin >> op;
    if (op == 1) {
      cin >> x;
      e[idx] = ++cur;
      if (h[x] == -1) {
        h[x] = idx;
        t[idx] = idx;
      } else {
        ne[t[h[x]]] = idx;
        t[h[x]] = idx;
      }
      ++idx;
    } else {
      cin >> x >> y;
      if (x == y || h[x] == -1)
        continue;
      else if (h[y] == -1)
        h[y] = h[x];
      else {
        ne[t[h[y]]] = h[x];
        t[h[y]] = t[h[x]];
      }
      h[x] = -1;
    }
  }
  for (int x = 1; x <= 500000; x++) {
    for (int i = h[x]; ~i; i = ne[i]) ans[e[i]] = x;
  }
  for (int i = 1; i <= idx; i++) cout << ans[i] << ' ';
  cout << '\n';
}

### Prompt

Your task is to create 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 = 5e5 + 5, INF = 0x3f3f3f3f;
int h[maxn], t[maxn], e[maxn], ne[maxn], cur, idx;
int ans[maxn];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  memset(h, -1, sizeof h);
  memset(ne, -1, sizeof ne);
  int n;
  cin >> n;
  while (n--) {
    int op, x, y;
    cin >> op;
    if (op == 1) {
      cin >> x;
      e[idx] = ++cur;
      if (h[x] == -1) {
        h[x] = idx;
        t[idx] = idx;
      } else {
        ne[t[h[x]]] = idx;
        t[h[x]] = idx;
      }
      ++idx;
    } else {
      cin >> x >> y;
      if (x == y || h[x] == -1)
        continue;
      else if (h[y] == -1)
        h[y] = h[x];
      else {
        ne[t[h[y]]] = h[x];
        t[h[y]] = t[h[x]];
      }
      h[x] = -1;
    }
  }
  for (int x = 1; x <= 500000; x++) {
    for (int i = h[x]; ~i; i = ne[i]) ans[e[i]] = x;
  }
  for (int i = 1; i <= idx; i++) cout << ans[i] << ' ';
  cout << '\n';
}

```