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;
int F() {
  char ch;
  int x, a;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-')
    ch = getchar(), a = -1;
  else
    a = 1;
  x = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0';
  return a * x;
}
int n;
int fa[200010], in[200010];
set<int> edge[200010];
int deg(int u) { return in[fa[u]] + ((int)(edge[u]).size()); }
int stk[200010], cnt;
int del[200010], tp;
void clear(int u) {
  stk[++cnt] = u;
  in[u] = 0;
  if (fa[u]) del[++tp] = u;
  for (auto v : edge[u]) clear(v);
}
void dfs(int u) {
  while (tp) {
    int x = del[tp--];
    if (fa[x]) edge[fa[x]].erase(x);
  }
  for (auto v : edge[u]) dfs(v);
  while (tp) {
    int x = del[tp--];
    if (fa[x]) edge[fa[x]].erase(x);
  }
  if (!in[u]) return;
  if (deg(u) & 1) return;
  clear(u);
}
int main() {
  ((n) = F());
  for (int i = 1; i <= n; ++i) {
    ((fa[i]) = F());
    in[i] = 1;
    if (fa[i]) edge[fa[i]].insert(i);
  }
  for (int i = 1; i <= n; ++i)
    if (!fa[i]) dfs(i);
  if (cnt != n) return puts("NO"), 0;
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d\n", stk[i]);
  return 0;
}