#include <bits/stdc++.h>
using namespace std;
int to[500000], of[500000], _index[500000];
vector<int> v;
int cnt;
int find(int i) {
  if (_index[i] == i) return i;
  return _index[i] = find(_index[i]);
}
void add(int x) {
  if (to[find(of[x])] != x) {
    cnt++;
    to[cnt] = x;
    of[x] = cnt;
    v.push_back(cnt);
  } else {
    v.push_back(of[x]);
  }
}
void change(int x, int y) {
  if (x == y) return;
  if (to[find(of[y])] != y) {
    cnt++;
    of[y] = cnt;
    to[cnt] = y;
  }
  int pa = find(of[x]), pb = find(of[y]);
  if (to[pa] != x) return;
  _index[pa] = pb;
}
void print() {
  for (auto x : v) cout << to[find(x)] << " ";
}
void solve() {
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) _index[i] = i;
  int t, x, y;
  cnt = 0;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      add(x);
    } else {
      cin >> x >> y;
      change(x, y);
    }
  }
  print();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
int to[500000], of[500000], _index[500000];
vector<int> v;
int cnt;
int find(int i) {
  if (_index[i] == i) return i;
  return _index[i] = find(_index[i]);
}
void add(int x) {
  if (to[find(of[x])] != x) {
    cnt++;
    to[cnt] = x;
    of[x] = cnt;
    v.push_back(cnt);
  } else {
    v.push_back(of[x]);
  }
}
void change(int x, int y) {
  if (x == y) return;
  if (to[find(of[y])] != y) {
    cnt++;
    of[y] = cnt;
    to[cnt] = y;
  }
  int pa = find(of[x]), pb = find(of[y]);
  if (to[pa] != x) return;
  _index[pa] = pb;
}
void print() {
  for (auto x : v) cout << to[find(x)] << " ";
}
void solve() {
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) _index[i] = i;
  int t, x, y;
  cnt = 0;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> x;
      add(x);
    } else {
      cin >> x >> y;
      change(x, y);
    }
  }
  print();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

```