#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  cin >> q;
  vector<long long> ret;
  vector<int> p(5e5 + 1, -1);
  function<int(int)> myFind = [&](int n) {
    if (p[n] == -1) {
      return n;
    }
    return p[n] = myFind(p[n]);
  };
  vector<tuple<long long, long long, long long> > query(q);
  for (int i = 0; i < q; ++i) {
    long long op, x, y;
    cin >> op;
    if (op == 1) {
      cin >> x;
    } else {
      cin >> x >> y;
    }
    query[i] = {op, x, y};
  }
  reverse((query).begin(), (query).end());
  map<long long, long long> m;
  for (auto [op, x, y] : query) {
    if (op == 1) {
      if (m.count(x)) {
        ret.push_back(m[x]);
      } else {
        ret.push_back(x);
      }
    } else {
      if (m.count(y)) {
        m[x] = m[y];
      } else {
        m[x] = y;
      }
    }
  }
  reverse((ret).begin(), (ret).end());
  for (auto e : ret) {
    cout << e << ' ';
  }
  cout << '\n';
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  cin >> q;
  vector<long long> ret;
  vector<int> p(5e5 + 1, -1);
  function<int(int)> myFind = [&](int n) {
    if (p[n] == -1) {
      return n;
    }
    return p[n] = myFind(p[n]);
  };
  vector<tuple<long long, long long, long long> > query(q);
  for (int i = 0; i < q; ++i) {
    long long op, x, y;
    cin >> op;
    if (op == 1) {
      cin >> x;
    } else {
      cin >> x >> y;
    }
    query[i] = {op, x, y};
  }
  reverse((query).begin(), (query).end());
  map<long long, long long> m;
  for (auto [op, x, y] : query) {
    if (op == 1) {
      if (m.count(x)) {
        ret.push_back(m[x]);
      } else {
        ret.push_back(x);
      }
    } else {
      if (m.count(y)) {
        m[x] = m[y];
      } else {
        m[x] = y;
      }
    }
  }
  reverse((ret).begin(), (ret).end());
  for (auto e : ret) {
    cout << e << ' ';
  }
  cout << '\n';
}

```