Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values p_i — the index of the router to which the i-th router was connected after being purchased (p_i < i).

There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n-th router.

Input

The first line contains integer number n (2 ≤ n ≤ 200000) — the number of the routers. The following line contains n-1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i is equal to index of the router to which the i-th was connected after purchase.

Output

Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.

Examples

Input

8
1 1 2 2 3 2 5


Output

1 2 5 8 

Input

6
1 2 3 4 5


Output

1 2 3 4 5 6 

Input

7
1 1 2 3 4 3


Output

1 3 7 
#include <bits/stdc++.h>
using namespace std;
const int M = 1e6 + 10;
struct tonod {
  int p, next;
} e[M];
int head[M], top = 1, fa[M];
void ins(int x, int y) {
  e[top].p = y, e[top].next = head[x], head[x] = top, top++;
}
void insert(int x, int y) { ins(x, y), ins(y, x); }
void dfs(int u, int f) {
  fa[u] = f;
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].p;
    if (v != f) dfs(v, u);
  }
}
int ans[M], cnt, a;
int main() {
  memset(head, -1, sizeof(head));
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d", &a), insert(a, i + 1);
  dfs(1, 0);
  while (fa[n]) ans[++cnt] = n, n = fa[n];
  printf("1 ");
  for (int i = cnt; i >= 1; i--) printf("%d ", ans[i]);
  printf("\n");
}
