#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long N = 5e5 + 5;
long long a[N], n;
vector<long long> ans;
vector<pair<long long, pair<long long, long long> > > q;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; ++i) a[i] = i;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long b;
    cin >> b;
    if (b == 1) {
      long long c;
      cin >> c;
      q.push_back({b, {c, -1}});
    } else {
      long long c, d;
      cin >> c >> d;
      q.push_back({b, {c, d}});
    }
  }
  reverse(q.begin(), q.end());
  for (long long i = 0; i < n; ++i) {
    if (q[i].first == 1) {
      ans.push_back(a[q[i].second.first]);
    } else {
      a[q[i].second.first] = a[q[i].second.second];
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto i : ans) cout << i << " ";
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long N = 5e5 + 5;
long long a[N], n;
vector<long long> ans;
vector<pair<long long, pair<long long, long long> > > q;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; ++i) a[i] = i;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long b;
    cin >> b;
    if (b == 1) {
      long long c;
      cin >> c;
      q.push_back({b, {c, -1}});
    } else {
      long long c, d;
      cin >> c >> d;
      q.push_back({b, {c, d}});
    }
  }
  reverse(q.begin(), q.end());
  for (long long i = 0; i < n; ++i) {
    if (q[i].first == 1) {
      ans.push_back(a[q[i].second.first]);
    } else {
      a[q[i].second.first] = a[q[i].second.second];
    }
  }
  reverse(ans.begin(), ans.end());
  for (auto i : ans) cout << i << " ";
  return 0;
}

```