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;
const int maxn = 5e5 + 4;
int ans[maxn], val[maxn], f[maxn], mp[maxn];
void init() {
  for (int i = 0; i < maxn; i++) f[i] = i;
}
int getv(int x) {
  if (f[x] == x) {
    ans[x] = val[x];
    return x;
  } else {
    f[x] = getv(f[x]);
    ans[x] = val[f[x]];
    return f[x];
  }
}
void solv() {
  init();
  int q, op, x, y, cnt = 0;
  cin >> q;
  for (int i = 0; i < q; i++) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d", &x);
      if (!mp[x]) mp[x] = cnt + 1;
      f[++cnt] = mp[x];
      val[mp[x]] = x;
    } else {
      scanf("%d %d", &x, &y);
      if (x == y) continue;
      if (!mp[y]) {
        val[f[mp[x]]] = y;
        mp[y] = mp[x];
        mp[x] = 0;
        continue;
      }
      f[mp[x]] = mp[y];
      mp[x] = 0;
    }
  }
  for (int i = 1; i <= cnt; i++) {
    ans[i] = val[getv(i)];
  }
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", ans[i]);
  }
}
signed main() { solv(); }
