#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
struct Dat {
  int f, x, y;
  Dat() {}
  Dat(int f, int x, int y) {
    this->f = f;
    this->x = x;
    this->y = y;
  }
};
Dat dat[N];
int fa[N];
int find(int x);
int ans[N];
int dfn[N];
int main() {
  int q;
  scanf("%d", &q);
  for (int i = 0; i < N; i++) {
    fa[i] = i;
    dfn[i] = N;
  }
  for (int i = 0; i < q; i++) {
    int f, x, y;
    scanf("%d%d", &f, &x);
    if (f == 2) {
      scanf("%d", &y);
    }
    dat[i] = Dat(f, x, y);
  }
  int cnt = 0;
  for (int i = q - 1; i >= 0; i--) {
    if (dat[i].f == 1) {
      ans[cnt++] = fa[dat[i].x];
    } else {
      fa[dat[i].x] = fa[dat[i].y];
    }
  }
  for (int i = cnt - 1; i >= 0; i--) {
    printf("%d%c", ans[i], i == 0 ? '\n' : ' ');
  }
  return 0;
}
int find(int x) {
  if (x == fa[x]) {
    return x;
  } else {
    if (dfn[fa[x]] > dfn[x]) {
      return fa[x] = find(fa[x]);
    }
    return fa[x];
  }
}

### Prompt

Please create a solution in cpp 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;
const int N = 1e6 + 5;
struct Dat {
  int f, x, y;
  Dat() {}
  Dat(int f, int x, int y) {
    this->f = f;
    this->x = x;
    this->y = y;
  }
};
Dat dat[N];
int fa[N];
int find(int x);
int ans[N];
int dfn[N];
int main() {
  int q;
  scanf("%d", &q);
  for (int i = 0; i < N; i++) {
    fa[i] = i;
    dfn[i] = N;
  }
  for (int i = 0; i < q; i++) {
    int f, x, y;
    scanf("%d%d", &f, &x);
    if (f == 2) {
      scanf("%d", &y);
    }
    dat[i] = Dat(f, x, y);
  }
  int cnt = 0;
  for (int i = q - 1; i >= 0; i--) {
    if (dat[i].f == 1) {
      ans[cnt++] = fa[dat[i].x];
    } else {
      fa[dat[i].x] = fa[dat[i].y];
    }
  }
  for (int i = cnt - 1; i >= 0; i--) {
    printf("%d%c", ans[i], i == 0 ? '\n' : ' ');
  }
  return 0;
}
int find(int x) {
  if (x == fa[x]) {
    return x;
  } else {
    if (dfn[fa[x]] > dfn[x]) {
      return fa[x] = find(fa[x]);
    }
    return fa[x];
  }
}

```