#include <bits/stdc++.h>
using namespace std;
const long long INF = 1E16;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int SIZE = 10000000;
long double _abs(long double a) {
  if (a < 0) return -a;
  return a;
}
long long _abss(long long a) {
  if (a < 0) return -a;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void solve() {
  long long k;
  cin >> k;
  vector<long long> a(k), x(k), y(k);
  for (long long i = 0; i < long long(k); ++i) {
    cin >> a[i];
    if (a[i] == 1) {
      cin >> x[i];
    } else {
      cin >> x[i] >> y[i];
    }
  }
  vector<long long> ans, p(SIZE, 0);
  for (long long i = 0; i < long long(SIZE); ++i) {
    p[i] = i;
  }
  for (long long i = long long(k) - 1; i >= 0; --i) {
    if (a[i] == 1) {
      ans.push_back(p[x[i]]);
    } else {
      p[x[i]] = p[y[i]];
    }
  }
  reverse((ans).begin(), (ans).end());
  for (auto i : ans) {
    cout << i << ' ';
  }
  cout << endl;
  return;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const long long INF = 1E16;
const long double EPS = 1E-9;
const long double PI = 3.1415926535897932384626433832795;
const int SIZE = 10000000;
long double _abs(long double a) {
  if (a < 0) return -a;
  return a;
}
long long _abss(long long a) {
  if (a < 0) return -a;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void solve() {
  long long k;
  cin >> k;
  vector<long long> a(k), x(k), y(k);
  for (long long i = 0; i < long long(k); ++i) {
    cin >> a[i];
    if (a[i] == 1) {
      cin >> x[i];
    } else {
      cin >> x[i] >> y[i];
    }
  }
  vector<long long> ans, p(SIZE, 0);
  for (long long i = 0; i < long long(SIZE); ++i) {
    p[i] = i;
  }
  for (long long i = long long(k) - 1; i >= 0; --i) {
    if (a[i] == 1) {
      ans.push_back(p[x[i]]);
    } else {
      p[x[i]] = p[y[i]];
    }
  }
  reverse((ans).begin(), (ans).end());
  for (auto i : ans) {
    cout << i << ' ';
  }
  cout << endl;
  return;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```