#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 5e6 + 7;
vector<vector<int>> q;
int ans[N];
int changeTo[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Q;
  cin >> Q;
  q.resize(Q);
  memset(changeTo, -1, sizeof changeTo);
  int n = 0;
  for (int i = 0; i < Q; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      n++;
      int x;
      cin >> x;
      q[i].push_back(1);
      q[i].push_back(x);
    } else {
      int x, y;
      cin >> x >> y;
      q[i].emplace_back(2);
      q[i].emplace_back(x);
      q[i].emplace_back(y);
    }
  }
  int idx = n - 1;
  for (int i = Q - 1; i >= 0; --i) {
    int t = q[i].front();
    if (t == 1) {
      int x = q[i].back();
      if (~changeTo[x]) {
        x = changeTo[x];
      }
      ans[idx--] = x;
    } else {
      int x = q[i][1];
      int y = q[i].back();
      int c = changeTo[y];
      if (~c)
        changeTo[x] = c;
      else
        changeTo[x] = y;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 5e6 + 7;
vector<vector<int>> q;
int ans[N];
int changeTo[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Q;
  cin >> Q;
  q.resize(Q);
  memset(changeTo, -1, sizeof changeTo);
  int n = 0;
  for (int i = 0; i < Q; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      n++;
      int x;
      cin >> x;
      q[i].push_back(1);
      q[i].push_back(x);
    } else {
      int x, y;
      cin >> x >> y;
      q[i].emplace_back(2);
      q[i].emplace_back(x);
      q[i].emplace_back(y);
    }
  }
  int idx = n - 1;
  for (int i = Q - 1; i >= 0; --i) {
    int t = q[i].front();
    if (t == 1) {
      int x = q[i].back();
      if (~changeTo[x]) {
        x = changeTo[x];
      }
      ans[idx--] = x;
    } else {
      int x = q[i][1];
      int y = q[i].back();
      int c = changeTo[y];
      if (~c)
        changeTo[x] = c;
      else
        changeTo[x] = y;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
  return 0;
}

```