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 
def bfs(start,end,g):
    vis = {start:-1}
    q = [start]
    found = False
    while q and not found:
        curr = q.pop(0)
        if curr == end:
            break
        for v in g[curr]:
            if v not in vis:
                vis[v] = curr
                q.append(v)
                if v == end:
                    found = True
                    break
    ans = []
    init = end
    while init != -1:
        ans.append(init)
        init = vis[init]
    return ans


n = int(input())
graph = [[] for _ in range(n)]
routers = [int(i) for i in input().split()]
for i in range(len(routers)):
    graph[routers[i]-1].append(i+1)
ans = bfs(0,n-1,graph)
i = 0
j = len(ans) - 1
while i <= j:
    ans[i],ans[j] = ans[j]+1,ans[i]+1
    i+=1
    j-=1
print(*ans)