#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long q;
    vector<vector<long long>> vec;
    vector<long long> actual(500007);
    for (long long i = 1; i <= 500005; i++) {
      actual[i] = i;
    }
    vector<long long> ans;
    cin >> q;
    long long n = q;
    while (q--) {
      long long x;
      cin >> x;
      if (x == 1) {
        long long y;
        cin >> y;
        vec.push_back({x, y});
      } else {
        long long y1, z1;
        cin >> y1 >> z1;
        vec.push_back({x, y1, z1});
      }
    }
    for (long long i = n - 1; i >= 0; i--) {
      if (vec[i][0] == 1) {
        ans.push_back(actual[vec[i][1]]);
      } else {
        actual[vec[i][1]] = actual[vec[i][2]];
      }
    }
    reverse(ans.begin(), ans.end());
    for (auto x : ans) {
      cout << x << " ";
    }
    cout << endl;
  }
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long q;
    vector<vector<long long>> vec;
    vector<long long> actual(500007);
    for (long long i = 1; i <= 500005; i++) {
      actual[i] = i;
    }
    vector<long long> ans;
    cin >> q;
    long long n = q;
    while (q--) {
      long long x;
      cin >> x;
      if (x == 1) {
        long long y;
        cin >> y;
        vec.push_back({x, y});
      } else {
        long long y1, z1;
        cin >> y1 >> z1;
        vec.push_back({x, y1, z1});
      }
    }
    for (long long i = n - 1; i >= 0; i--) {
      if (vec[i][0] == 1) {
        ans.push_back(actual[vec[i][1]]);
      } else {
        actual[vec[i][1]] = actual[vec[i][2]];
      }
    }
    reverse(ans.begin(), ans.end());
    for (auto x : ans) {
      cout << x << " ";
    }
    cout << endl;
  }
}

```