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>
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, p[N], la[N], ne[N], rt, d[N];
int an[N], ans, f[N][3], ct[N][3], b[N];
void dfs(int x) {
  for (int i = la[x]; i; i = ne[i]) dfs(i);
  int cnt = 0, cnt2 = 0, cnt3 = 0, cnt4 = 0;
  for (int i = la[x]; i; i = ne[i]) {
    cnt += f[i][2];
    if (f[i][2])
      cnt2 += f[i][1];
    else
      cnt3 += f[i][1];
  }
  ct[x][1] = cnt;
  ct[x][2] = cnt2;
  ct[x][3] = cnt3;
  if ((!((d[x] - cnt) & 1) || cnt2 > 0) && (cnt3 + cnt == d[x])) f[x][1] = 1;
  if ((((d[x] - cnt) & 1) || cnt2 > 0) && (cnt3 + cnt == d[x])) f[x][2] = 1;
}
void fi(int x, int st) {
  int ne1;
  b[x] = st;
  if (st == 1)
    ne1 = !((d[x] - ct[x][1]) & 1) && ct[x][2] > 0;
  else
    ne1 = ((d[x] - ct[x][1]) & 1) && ct[x][2] > 0;
  for (int i = la[x]; i; i = ne[i]) {
    if (f[i][2]) {
      fi(i, 2);
      if (ne1 && f[i][1]) ne1--, fi(i, 1);
    }
  }
  an[++ans] = x;
  for (int i = la[x]; i; i = ne[i])
    if (!b[i]) fi(i, 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, E = n; i <= E; i++)
    scanf("%d", &p[i]),
        (p[i] ? ne[i] = la[p[i]], la[p[i]] = i, d[p[i]]++ : rt = i);
  dfs(rt);
  if (f[rt][1]) {
    printf("YES\n");
    fi(rt, 1);
    for (int i = 1, E = n; i <= E; i++) printf("%d ", an[i]);
  } else
    printf("NO\n");
}
