#include <bits/stdc++.h>
using namespace std;
const long long int mma = 1e6 + 10;
const long long int mod = 1e9 + 7;
long long int par[mma];
vector<long long int> g[mma];
long long int get_par(long long int x) {
  if (x == par[x]) {
    return x;
  }
  return par[x] = get_par(par[x]);
}
void solve() {
  long long int q;
  cin >> q;
  for (int i = 0; i < mma; i++) {
    par[i] = i;
  }
  long long int k = 0;
  for (int i = 0; i < q; i++) {
    long long int u;
    cin >> u;
    if (u == 1) {
      long long int x;
      cin >> x;
      g[i].push_back(x);
      k++;
    } else {
      long long int x, y;
      cin >> x >> y;
      g[i].push_back(x);
      g[i].push_back(y);
    }
  }
  long long int a[k];
  long long int u = 0;
  for (int i = q - 1; i >= 0; i--) {
    if (g[i].size() == 2) {
      par[g[i][0]] = par[g[i][1]];
    } else {
      a[k - u - 1] = par[g[i][0]];
      u++;
    }
  }
  for (int i = 0; i < k; i++) {
    cout << a[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### 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 long long int mma = 1e6 + 10;
const long long int mod = 1e9 + 7;
long long int par[mma];
vector<long long int> g[mma];
long long int get_par(long long int x) {
  if (x == par[x]) {
    return x;
  }
  return par[x] = get_par(par[x]);
}
void solve() {
  long long int q;
  cin >> q;
  for (int i = 0; i < mma; i++) {
    par[i] = i;
  }
  long long int k = 0;
  for (int i = 0; i < q; i++) {
    long long int u;
    cin >> u;
    if (u == 1) {
      long long int x;
      cin >> x;
      g[i].push_back(x);
      k++;
    } else {
      long long int x, y;
      cin >> x >> y;
      g[i].push_back(x);
      g[i].push_back(y);
    }
  }
  long long int a[k];
  long long int u = 0;
  for (int i = q - 1; i >= 0; i--) {
    if (g[i].size() == 2) {
      par[g[i][0]] = par[g[i][1]];
    } else {
      a[k - u - 1] = par[g[i][0]];
      u++;
    }
  }
  for (int i = 0; i < k; i++) {
    cout << a[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```