#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int ans = power(a, b / 2);
  if (b % 2 == 0) return (ans * ans) % 1000000007;
  return ((ans * ans) % 1000000007 * a) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q, p = 0, i;
  cin >> q;
  map<int, vector<int>> m;
  while (q--) {
    int f, x, y;
    cin >> f;
    if (f == 1) {
      cin >> x;
      p++;
      m[x].push_back(p);
    } else {
      cin >> x >> y;
      if (x != y) {
        if (m[x].size() > m[y].size()) m[x].swap(m[y]);
        vector<int> vt;
        vt = m[x];
        for (i = 0; i < vt.size(); i++) {
          m[y].push_back(vt[i]);
        }
        m.erase(x);
      }
    }
  }
  vector<pair<int, int>> v;
  for (auto it = m.begin(); it != m.end(); it++) {
    vector<int> vt;
    int r = it->first;
    vt = m[r];
    for (i = 0; i < vt.size(); i++) {
      v.push_back(make_pair(vt[i], r));
    }
  }
  sort(v.begin(), v.end());
  for (i = 0; i < v.size(); i++) {
    cout << v[i].second << " ";
  }
  cout << "\n";
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int ans = power(a, b / 2);
  if (b % 2 == 0) return (ans * ans) % 1000000007;
  return ((ans * ans) % 1000000007 * a) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q, p = 0, i;
  cin >> q;
  map<int, vector<int>> m;
  while (q--) {
    int f, x, y;
    cin >> f;
    if (f == 1) {
      cin >> x;
      p++;
      m[x].push_back(p);
    } else {
      cin >> x >> y;
      if (x != y) {
        if (m[x].size() > m[y].size()) m[x].swap(m[y]);
        vector<int> vt;
        vt = m[x];
        for (i = 0; i < vt.size(); i++) {
          m[y].push_back(vt[i]);
        }
        m.erase(x);
      }
    }
  }
  vector<pair<int, int>> v;
  for (auto it = m.begin(); it != m.end(); it++) {
    vector<int> vt;
    int r = it->first;
    vt = m[r];
    for (i = 0; i < vt.size(); i++) {
      v.push_back(make_pair(vt[i], r));
    }
  }
  sort(v.begin(), v.end());
  for (i = 0; i < v.size(); i++) {
    cout << v[i].second << " ";
  }
  cout << "\n";
  return 0;
}

```