#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long int n = ({
    long long int tito;
    cin >> tito;
    tito;
  });
  vector<vector<long long int>> v;
  map<long long int, vector<long long int>> mm;
  long long int pos = 0;
  for (long long int i = 0; i < n; i++) {
    long long int a = ({
      long long int tito;
      cin >> tito;
      tito;
    });
    long long int b, c;
    if (a == 1) {
      b = ({
        long long int tito;
        cin >> tito;
        tito;
      });
      c = 0;
    } else {
      b = ({
        long long int tito;
        cin >> tito;
        tito;
      });
      c = ({
        long long int tito;
        cin >> tito;
        tito;
      });
    }
    vector<long long int> vv;
    vv.push_back(a);
    vv.push_back(b);
    vv.push_back(c);
    v.push_back(vv);
  }
  vector<long long int> ans;
  vector<long long int> aa;
  for (long long int i = 0; i < 5 * 1e5 + 5; i++) {
    aa.push_back(i);
  }
  for (long long int i = 0; i < n; i++) {
    vector<long long int> vv = v[n - 1 - i];
    long long int a = vv[0];
    long long int b = vv[1];
    long long int c = vv[2];
    if (a == 1) {
      ans.push_back(aa[b]);
    } else {
      aa[b] = aa[c];
    }
  }
  long long int nn = ans.size();
  for (long long int i = 0; i < nn; i++) {
    cout << ans[nn - 1 - i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
void solve() {
  long long int n = ({
    long long int tito;
    cin >> tito;
    tito;
  });
  vector<vector<long long int>> v;
  map<long long int, vector<long long int>> mm;
  long long int pos = 0;
  for (long long int i = 0; i < n; i++) {
    long long int a = ({
      long long int tito;
      cin >> tito;
      tito;
    });
    long long int b, c;
    if (a == 1) {
      b = ({
        long long int tito;
        cin >> tito;
        tito;
      });
      c = 0;
    } else {
      b = ({
        long long int tito;
        cin >> tito;
        tito;
      });
      c = ({
        long long int tito;
        cin >> tito;
        tito;
      });
    }
    vector<long long int> vv;
    vv.push_back(a);
    vv.push_back(b);
    vv.push_back(c);
    v.push_back(vv);
  }
  vector<long long int> ans;
  vector<long long int> aa;
  for (long long int i = 0; i < 5 * 1e5 + 5; i++) {
    aa.push_back(i);
  }
  for (long long int i = 0; i < n; i++) {
    vector<long long int> vv = v[n - 1 - i];
    long long int a = vv[0];
    long long int b = vv[1];
    long long int c = vv[2];
    if (a == 1) {
      ans.push_back(aa[b]);
    } else {
      aa[b] = aa[c];
    }
  }
  long long int nn = ans.size();
  for (long long int i = 0; i < nn; i++) {
    cout << ans[nn - 1 - i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) solve();
  return 0;
}

```