#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 99;
int fu[2 * N];
map<int, int> mapa;
map<int, int> rmapa;
void dodaj(int i) {
  int x;
  cin >> x;
  if (mapa[x] == 0 || fu[mapa[x]] != x) {
    mapa[x] = N + i;
    fu[N + i] = x;
  }
  fu[i] = mapa[x];
}
void zamien() {
  int x, y;
  cin >> x >> y;
  if (x == y) return;
  if (fu[mapa[x]] == x) {
    if (fu[mapa[y]] == y)
      fu[mapa[x]] = mapa[y];
    else {
      mapa[y] = mapa[x];
      fu[mapa[x]] = y;
    }
  }
}
int findd(int i) {
  if (fu[i] < N) return fu[i];
  fu[i] = findd(fu[i]);
  return fu[i];
}
int main() {
  int K = 0;
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int a;
    cin >> a;
    if (a == 1) {
      dodaj(K);
      K++;
    }
    if (a == 2) zamien();
  }
  for (int i = 0; i < K; i++) {
    cout << findd(i) << " ";
  }
  cout << "\n";
}

### 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;
const int N = 5e5 + 99;
int fu[2 * N];
map<int, int> mapa;
map<int, int> rmapa;
void dodaj(int i) {
  int x;
  cin >> x;
  if (mapa[x] == 0 || fu[mapa[x]] != x) {
    mapa[x] = N + i;
    fu[N + i] = x;
  }
  fu[i] = mapa[x];
}
void zamien() {
  int x, y;
  cin >> x >> y;
  if (x == y) return;
  if (fu[mapa[x]] == x) {
    if (fu[mapa[y]] == y)
      fu[mapa[x]] = mapa[y];
    else {
      mapa[y] = mapa[x];
      fu[mapa[x]] = y;
    }
  }
}
int findd(int i) {
  if (fu[i] < N) return fu[i];
  fu[i] = findd(fu[i]);
  return fu[i];
}
int main() {
  int K = 0;
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int a;
    cin >> a;
    if (a == 1) {
      dodaj(K);
      K++;
    }
    if (a == 2) zamien();
  }
  for (int i = 0; i < K; i++) {
    cout << findd(i) << " ";
  }
  cout << "\n";
}

```