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;
const int NMAX = 5e5;
struct query {
  int type;
  int x;
  int y;
} q[NMAX];
vector<int> v;
int fin[NMAX + 1];
void solve() {
  int n, i, a, b, c;
  cin >> n;
  for (i = 0; i <= NMAX; i++) fin[i] = i;
  for (i = 0; i < n; i++) {
    cin >> q[i].type;
    if (q[i].type == 1) {
      cin >> q[i].x;
    } else
      cin >> q[i].x >> q[i].y;
  }
  for (i = n - 1; i >= 0; i--) {
    if (q[i].type == 1) {
      v.push_back(fin[q[i].x]);
    } else
      fin[q[i].x] = fin[q[i].y];
  }
  reverse(v.begin(), v.end());
  for (auto it : v) cout << it << ' ';
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
