#include <bits/stdc++.h>
using namespace std;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
int dp[500050];
pair<int, int> a[500050];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  for (int i = 1, op; i <= q; ++i) {
    cin >> op;
    if (op == 1) {
      int x;
      cin >> x;
      a[i] = {x, -1};
    } else {
      int x, y;
      cin >> x >> y;
      a[i] = {x, y};
    }
  }
  vector<int> ans;
  for (int i = q; i >= 1; --i) {
    if (a[i].second == -1) {
      if (dp[a[i].first] == 0)
        ans.push_back(a[i].first), dp[a[i].first] = a[i].first;
      else
        ans.push_back(dp[a[i].first]);
    } else {
      int x = a[i].first, y = a[i].second;
      if (dp[y] == 0) dp[y] = y;
      dp[x] = dp[y];
    }
  }
  reverse(ans.begin(), ans.end());
  print(ans);
}

### Prompt

Please create a solution in cpp 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;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
int dp[500050];
pair<int, int> a[500050];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  for (int i = 1, op; i <= q; ++i) {
    cin >> op;
    if (op == 1) {
      int x;
      cin >> x;
      a[i] = {x, -1};
    } else {
      int x, y;
      cin >> x >> y;
      a[i] = {x, y};
    }
  }
  vector<int> ans;
  for (int i = q; i >= 1; --i) {
    if (a[i].second == -1) {
      if (dp[a[i].first] == 0)
        ans.push_back(a[i].first), dp[a[i].first] = a[i].first;
      else
        ans.push_back(dp[a[i].first]);
    } else {
      int x = a[i].first, y = a[i].second;
      if (dp[y] == 0) dp[y] = y;
      dp[x] = dp[y];
    }
  }
  reverse(ans.begin(), ans.end());
  print(ans);
}

```