Problem: 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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> G[200100];
int n, dp[200100], anses[200100], tp;
void dfs(int u, int f) {
  for (auto p : G[u])
    if (p != f) dfs(p, u);
  if (G[u].size() == 1) return (void)(dp[u] = 0);
  int cnt = 0;
  for (auto p : G[u])
    if (p != f && dp[p] == 0) cnt++;
  dp[u] = cnt & 1;
  if (dp[u]) anses[++tp] = u;
  if (f == 0 && dp[u] == 1) {
    printf("NO\n");
    exit(0);
  }
}
void dfs2(int u, int f) {
  if (!dp[u]) anses[++tp] = u;
  for (auto p : G[u])
    if (p != f) dfs2(p, u);
}
int main() {
  scanf("%d", &n);
  if (n == 1) return printf("YES\n1"), 0;
  if (n == 2) return printf("NO"), 0;
  for (int i = 1, u; i <= n; ++i) {
    scanf("%d", &u);
    if (u) G[u].push_back(i), G[i].push_back(u);
  }
  int rt = 0;
  for (int i = 1; i <= n; ++i)
    if (G[i].size() > 1) rt = i;
  dfs(rt, 0), dfs2(rt, 0);
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d\n", anses[i]);
}