#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  map<long long, long long> m;
  for (int i = 1; i <= 5 * pow(10, 5); i++) {
    m[i] = i;
  }
  long long q;
  cin >> q;
  vector<pair<long long, pair<long long, long long>>> v;
  long long n = 0;
  for (int i = 0; i < q; i++) {
    long long inp1;
    cin >> inp1;
    if (inp1 == 1) {
      n++;
      long long inp2, inp3;
      cin >> inp2;
      inp3 = inp2;
      v.push_back({inp1, {inp2, inp3}});
    } else {
      long long inp2, inp3;
      cin >> inp2 >> inp3;
      v.push_back({inp1, {inp2, inp3}});
    }
  }
  long long ans[n];
  long long m1 = n;
  n--;
  set<long long> st;
  for (int i = q - 1; i >= 0; i--) {
    if (v[i].first == 1) {
      ans[n] = m[v[i].second.first];
      n--;
    } else {
      if (st.find(v[i].second.second) != st.end()) {
        m[v[i].second.first] = m[v[i].second.second];
        st.insert(v[i].second.first);
        continue;
      } else {
        m[v[i].second.first] = m[v[i].second.second];
        st.insert(v[i].second.first);
      }
    }
  }
  for (int i = 0; i < m1; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  map<long long, long long> m;
  for (int i = 1; i <= 5 * pow(10, 5); i++) {
    m[i] = i;
  }
  long long q;
  cin >> q;
  vector<pair<long long, pair<long long, long long>>> v;
  long long n = 0;
  for (int i = 0; i < q; i++) {
    long long inp1;
    cin >> inp1;
    if (inp1 == 1) {
      n++;
      long long inp2, inp3;
      cin >> inp2;
      inp3 = inp2;
      v.push_back({inp1, {inp2, inp3}});
    } else {
      long long inp2, inp3;
      cin >> inp2 >> inp3;
      v.push_back({inp1, {inp2, inp3}});
    }
  }
  long long ans[n];
  long long m1 = n;
  n--;
  set<long long> st;
  for (int i = q - 1; i >= 0; i--) {
    if (v[i].first == 1) {
      ans[n] = m[v[i].second.first];
      n--;
    } else {
      if (st.find(v[i].second.second) != st.end()) {
        m[v[i].second.first] = m[v[i].second.second];
        st.insert(v[i].second.first);
        continue;
      } else {
        m[v[i].second.first] = m[v[i].second.second];
        st.insert(v[i].second.first);
      }
    }
  }
  for (int i = 0; i < m1; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

```