#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = (int)0x3f3f3f3f3f3f3f3f, ninf = (int)0xc0c0c0c0c0c0c0c0;
using ll = long long;
using ld = long double;
void solve() {
  const int N = (int)5e5 + 3;
  int q;
  cin >> q;
  vector<vector<int> > color(N);
  int id = 0;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      color[x].push_back(++id);
    } else {
      int x, y;
      cin >> x >> y;
      if (x == y) continue;
      if (((int)(color[y]).size()) < ((int)(color[x]).size()))
        swap(color[x], color[y]);
      color[y].insert(color[y].end(), color[x].begin(), color[x].end());
      color[x].clear();
    }
  }
  vector<int> ans(N);
  for (int i = 1; i <= N; ++i) {
    for (auto& x : color[i]) {
      ans[x] = i;
    }
  }
  for (int i = 1; i <= id; ++i) cout << ans[i] << ' ';
  cout << '\n';
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = (int)0x3f3f3f3f3f3f3f3f, ninf = (int)0xc0c0c0c0c0c0c0c0;
using ll = long long;
using ld = long double;
void solve() {
  const int N = (int)5e5 + 3;
  int q;
  cin >> q;
  vector<vector<int> > color(N);
  int id = 0;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      color[x].push_back(++id);
    } else {
      int x, y;
      cin >> x >> y;
      if (x == y) continue;
      if (((int)(color[y]).size()) < ((int)(color[x]).size()))
        swap(color[x], color[y]);
      color[y].insert(color[y].end(), color[x].begin(), color[x].end());
      color[x].clear();
    }
  }
  vector<int> ans(N);
  for (int i = 1; i <= N; ++i) {
    for (auto& x : color[i]) {
      ans[x] = i;
    }
  }
  for (int i = 1; i <= id; ++i) cout << ans[i] << ' ';
  cout << '\n';
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```