Destruction of a Tree (贪心)

You are given a tree (a graph with n vertices and n - 1 edges in which it's possible to reach any vertex from any other vertex using only its edges).

A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.

Destroy all vertices in the given tree or determine that it is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 2·10⁵) — number of vertices in a tree.

The second line contains n integers p₁, p₂, ..., p_n (0 ≤ p_i ≤ n). If p_i ≠ 0 there is an edge between vertices i and p_i. It is guaranteed that the given graph is a tree.

Output

If it's possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).

If it's possible to destroy all vertices, in the next n lines print the indices of the vertices in order you destroy them. If there are multiple correct answers, print any.

Examples

Input

5
0 1 2 1 2

Output

YES
1
2
3
5
4

Input

4
0 1 2 3

Output

NO

Note

In the first example at first you have to remove the vertex with index 1 (after that, the edges (1, 2) and (1, 4) are removed), then the vertex with index 2 (and edges (2, 3) and (2, 5) are removed). After that there are no edges in the tree, so you can remove remaining vertices in any order.

注意到每次删点都是删去偶数条边(只能删去度数为偶数的点）, 当n为偶数时无解。

考虑从根开始dfs，然后从底向上处理结点，当处理到结点u时,子树要么为空，要么为所有点度数为奇数的树。若u度数为偶数，以u为根的树可以依次处理，删去。

#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 2e5+10;
vector<int>G[N];
int vis[N],rt=0;
void print(int x){
	vis[x]=1;
	printf("%d\n",x);
	for(int i=0;i<G[x].size();i++){
		int v=G[x][i];
		if(!vis[v]) print(v);
	} 
}
void DFS(int u,int fa){
	int d=0;
	for(int i=0;i<G[u].size();i++){
		int v=G[u][i];
		if(!vis[v]) DFS(v,u);
	}
	for(int i=0;i<G[u].size();i++){
		int v=G[u][i];
		if(!vis[v]) d++;
	}
	if(fa) d++;
	if(d%2==0){
		printf("%d\n",u);
		vis[u]=1;
		for(int i=0;i<G[u].size();i++){
			int v=G[u][i];
			if(!vis[v]) print(v);
		}
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		int fa;
		scanf("%d",&fa);
		if(!fa) rt=i;
		else G[fa].push_back(i);
	}	
	if((n&1)==0) printf("NO\n");
	else{
		printf("YES\n");
		DFS(rt,0);	
	}
	return 0;
}