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>
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
void read(int &x) {
  char c = getchar();
  int k = 1;
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  for (x = 0; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= k;
}
const int N = 200005;
int n, m, x, y, s, k, ans, f[N];
struct edge {
  int fa;
} a[N];
void dfs(int k) {
  f[++m] = k;
  if (k == 1) {
    return;
  }
  dfs(a[k].fa);
}
int main() {
  read(n);
  for (int i = 2; i <= n; i++) {
    read(x);
    a[i].fa = x;
  }
  dfs(n);
  for (int i = m; i; --i) {
    printf("%d ", f[i]);
  }
  return 0;
}
