#include <bits/stdc++.h>
using namespace std;
template <typename _Tp>
bool minimize(_Tp &__a, const _Tp &__b) {
  if (__a > __b) {
    __a = __b;
    return true;
  }
  return false;
}
template <typename _Tp>
bool maximize(_Tp &__a, const _Tp &__b) {
  if (__a < __b) {
    __a = __b;
    return true;
  }
  return false;
}
const int siz = 1e5 + 2;
const int SIZ = 5e5 + 2;
const int mod = 1e9 + 7;
const int maxx = 2e9;
const long long MAXX = 1e18;
const string file = "1";
int val[SIZ];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  vector<vector<int> > queries;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      queries.push_back({1, x});
    } else {
      int x, y;
      cin >> x >> y;
      queries.push_back({2, x, y});
    }
  }
  reverse((queries).begin(), (queries).end());
  for (int i = (int)(1); i <= (int)(SIZ - 1); i++) {
    val[i] = i;
  }
  vector<int> ans;
  for (auto qr : queries) {
    int t = qr[0];
    if (t == 1) {
      int x = qr[1];
      ans.push_back(val[x]);
    } else {
      int x = qr[1], y = qr[2];
      val[x] = val[y];
    }
  }
  reverse((ans).begin(), (ans).end());
  for (auto x : ans) {
    cout << x << " ";
  }
  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;
template <typename _Tp>
bool minimize(_Tp &__a, const _Tp &__b) {
  if (__a > __b) {
    __a = __b;
    return true;
  }
  return false;
}
template <typename _Tp>
bool maximize(_Tp &__a, const _Tp &__b) {
  if (__a < __b) {
    __a = __b;
    return true;
  }
  return false;
}
const int siz = 1e5 + 2;
const int SIZ = 5e5 + 2;
const int mod = 1e9 + 7;
const int maxx = 2e9;
const long long MAXX = 1e18;
const string file = "1";
int val[SIZ];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  vector<vector<int> > queries;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      queries.push_back({1, x});
    } else {
      int x, y;
      cin >> x >> y;
      queries.push_back({2, x, y});
    }
  }
  reverse((queries).begin(), (queries).end());
  for (int i = (int)(1); i <= (int)(SIZ - 1); i++) {
    val[i] = i;
  }
  vector<int> ans;
  for (auto qr : queries) {
    int t = qr[0];
    if (t == 1) {
      int x = qr[1];
      ans.push_back(val[x]);
    } else {
      int x = qr[1], y = qr[2];
      val[x] = val[y];
    }
  }
  reverse((ans).begin(), (ans).end());
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << "\n";
  return 0;
}

```