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, ..., pn (0 ≤ pi ≤ n). If pi ≠ 0 there is an edge between vertices i and pi. 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.

<image>
import java.io.*;
import java.util.*;

public class DestructionOfTree {
	
	static ArrayList<Integer>[] e;
	static boolean[] relaxed, vis;
	static PrintWriter pw;
	public static void main(String[] args) throws Exception{
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		
		int n = Integer.parseInt(f.readLine());
		
		if(n % 2 == 0) {
			pw.println("NO");
			pw.close();
			System.exit(0);
		}
		
		pw.println("YES");
		e = new ArrayList[n];
		relaxed = new boolean[n];
		vis = new boolean[n];
		for(int i = 0;i<n;i++) {
			e[i] = new ArrayList<Integer>();
		}
		
		StringTokenizer st = new StringTokenizer(f.readLine());
		for(int i = 1;i<=n;i++) {
			int x = Integer.parseInt(st.nextToken());
			
			if(x != 0) {
				e[i - 1].add(x-1);
				e[x - 1].add(i-1);
			}
		}
		
		dfs(0, -1);
		
		print(0);
		
		pw.close();
	}
	
	static boolean dfs(int x, int p) {
		
		if(e[x].size() == 1 && e[x].get(0) == p) {
			return false;
		}
		
		int counter = 0;
		for(int n: e[x]) {
			if(n != p) {
				if(!dfs(n,x)) {
					counter ++;
				}
			}
		}
		if((x == 0 && counter % 2 == 0) || (x != 0 && counter % 2 == 1)) {
			pw.println(x+1);
			relaxed[x] = true;
			return true;
		}
		else {
			return false;
		}
	}
	
	static void print(int x) {
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(x);
		if(!relaxed[x]) pw.println(x+1);
		vis[x] = true;
		
		while(!q.isEmpty()) {
			int y = q.poll();
			
			for(int n: e[y]) {
				if(!vis[n]) {
					if(!relaxed[n]) pw.println(n+1);
					q.add(n);
					vis[n] = true;
				}
			}
		}
	}
	

}
