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 = 200010;
int k, n, m, x, Rt;
int h[N], nx[N << 1], t[N << 1], num;
int A[N], B[N], C[N], l1, l2, l3;
int a[N], cnt;
bool f[N][2], fl;
bool g[2][N], d[N];
set<int> S;
void Add(int x, int y) {
  t[++num] = y;
  nx[num] = h[x];
  h[x] = num;
}
void Dfs(int x) {
  for (int i = h[x]; i; i = nx[i]) Dfs(t[i]);
  l1 = l2 = l3 = 0;
  for (int i = h[x]; i; i = nx[i]) {
    if (f[t[i]][0] && f[t[i]][1])
      A[++l1] = t[i];
    else if (f[t[i]][0])
      B[++l2] = t[i];
    else if (f[t[i]][1])
      C[++l3] = t[i];
  }
  if (!l1) {
    bool t = d[x] ^ (l2 & 1);
    f[x][t] = 1;
    for (int i = 1; i <= l3; i++) g[t][C[i]] = 1;
  } else {
    f[x][0] = f[x][1] = 1;
    for (int t = 0; t < 2; t++) {
      if (!(t ^ (l2 & 1) ^ d[x])) g[t][A[1]] = 1;
      for (int i = 2; i <= l1; i++) g[t][A[i]] = 1;
      for (int i = 1; i <= l3; i++) g[t][C[i]] = 1;
    }
  }
  if (!f[x][0] && !f[x][1]) fl = 1;
}
void Get(int x, bool d) {
  for (int i = h[x]; i; i = nx[i])
    if (!g[d][t[i]]) Get(t[i], 0);
  a[++cnt] = x;
  for (int i = h[x]; i; i = nx[i])
    if (g[d][t[i]]) Get(t[i], 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x)
      Add(x, i), d[x] ^= 1, d[i] ^= 1;
    else
      Rt = i;
  }
  Dfs(Rt);
  if (fl || !f[Rt][0])
    puts("NO");
  else {
    puts("YES");
    Get(Rt, 0);
    for (int i = 1; i <= n; i++) printf("%d\n", a[i]);
  }
  return 0;
}
