#include <bits/stdc++.h>
const int N = 5e5;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int my_rand(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
int par[N + 5];
void init() {
  for (long long int i = 1; i <= N; i++) par[i] = i;
}
int get(int n) {
  if (par[n] == n) return n;
  return par[n] = get(par[n]);
}
void uni(int a, int b) { par[a] = b; }
int v[500005];
int in[500005];
void solve() {
  init();
  long long int nw = 1e5;
  long long int q;
  long long int i = 1;
  cin >> q;
  int x, y;
  while (q--) {
    long long int t;
    cin >> t;
    if (t == 1) {
      cin >> x;
      v[i] = x;
      if (in[x]) {
        uni(get(in[x]), i);
      } else {
        in[x] = i;
      }
      i++;
    } else {
      cin >> x >> y;
      int xin = in[x];
      if (xin == 0) continue;
      long long int yin = in[y];
      yin = get(yin);
      xin = get(xin);
      if (yin == 0) {
        yin = xin;
        v[yin] = y;
        in[y] = yin;
        in[x] = 0;
      } else {
        if (get(in[x]) == get(in[y])) continue;
        uni(get(in[x]), get(in[y]));
        in[x] = 0;
      }
    }
  }
  for (long long int j = 1; j <= i - 1; j++) {
    x = get(j);
    cout << v[x] << ' ';
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  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>
const int N = 5e5;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int my_rand(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
int par[N + 5];
void init() {
  for (long long int i = 1; i <= N; i++) par[i] = i;
}
int get(int n) {
  if (par[n] == n) return n;
  return par[n] = get(par[n]);
}
void uni(int a, int b) { par[a] = b; }
int v[500005];
int in[500005];
void solve() {
  init();
  long long int nw = 1e5;
  long long int q;
  long long int i = 1;
  cin >> q;
  int x, y;
  while (q--) {
    long long int t;
    cin >> t;
    if (t == 1) {
      cin >> x;
      v[i] = x;
      if (in[x]) {
        uni(get(in[x]), i);
      } else {
        in[x] = i;
      }
      i++;
    } else {
      cin >> x >> y;
      int xin = in[x];
      if (xin == 0) continue;
      long long int yin = in[y];
      yin = get(yin);
      xin = get(xin);
      if (yin == 0) {
        yin = xin;
        v[yin] = y;
        in[y] = yin;
        in[x] = 0;
      } else {
        if (get(in[x]) == get(in[y])) continue;
        uni(get(in[x]), get(in[y]));
        in[x] = 0;
      }
    }
  }
  for (long long int j = 1; j <= i - 1; j++) {
    x = get(j);
    cout << v[x] << ' ';
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

```