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].
1
1 50
 1
1 114514
 1
1 2002
 1
1 500

50 
 114514 
 2002 
 500 

#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = (int)0x3f3f3f3f3f3f3f3f, ninf = (int)0xc0c0c0c0c0c0c0c0;
using ll = long long;
using ld = long double;
void solve() {
  const int N = (int)5e5 + 3;
  vector<int> prev(N), color(N);
  vector<int> p(N, -1);
  function<int(int)> root = [&](int u) -> int {
    return p[u] < 0 ? u : p[u] = root(p[u]);
  };
  function<bool(int, int, int)> unite = [&](int u, int v, int c) -> bool {
    if ((u = root(u)) == (v = root(v))) return 0;
    if (p[u] > p[v]) swap(u, v);
    color[u] = c;
    color[v] = 0;
    p[u] += p[v];
    p[v] = u;
    return 1;
  };
  int q;
  cin >> q;
  int id = 0;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int x;
      cin >> x;
      ++id;
      if (prev[x]) {
        unite(id, prev[x], x);
      } else {
        prev[x] = id;
        color[root(id)] = x;
      }
    } else {
      int x, y;
      cin >> x >> y;
      if (!prev[x]) continue;
      if (prev[y]) {
        unite(prev[y], prev[x], y);
        prev[x] = 0;
      } else {
        color[root(prev[x])] = y;
        prev[y] = prev[x];
        prev[x] = 0;
      }
    }
  }
  for (int i = 1; i <= id; ++i) cout << color[root(i)] << ' ';
  cout << '\n';
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
