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].
1
1 50
 1
1 114514
 1
1 2002
 1
1 500

50 
 114514 
 2002 
 500 

#include <bits/stdc++.h>
using namespace std;
void init_code() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  init_code();
  vector<long long int> v;
  map<pair<long long int, long long int>, long long int> m;
  long long int TT;
  long long int count = 0;
  cin >> TT;
  for (long long int t = 0; t < TT; t++) {
    long long int n;
    cin >> n;
    if (n == 1) {
      long long int tt;
      cin >> tt;
      v.push_back(tt);
      count++;
    } else {
      long long int t1, t2;
      cin >> t1 >> t2;
      m.insert({{t1, count}, t2});
    }
  }
  for (auto x : m) {
    if (x.first.second == 0) {
      continue;
    } else {
      for (long long int j = 0; j < x.first.second; j++) {
        if (v[j] == x.first.first) {
          v[j] = x.second;
        }
      }
    }
  };
  for (auto n : v) {
    cout << n << " ";
  }
  cout << "\n";
  ;
  return 0;
}
