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;
template <typename T>
inline void gmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
inline void gmin(T &x, T y) {
  x = x < y ? x : y;
}
const int MAXN = 200010;
int n, rt;
int f[MAXN];
int dp[MAXN];
vector<int> G[MAXN];
void dfs(int x) {
  int sz = G[x].size();
  for (register unsigned i = 0; i < G[x].size(); i++) {
    int y = G[x][i];
    dfs(y);
    if (dp[y]) sz--;
  }
  dp[x] = sz & 1;
}
void dfs2(int x) {
  for (register unsigned i = 0; i < G[x].size(); i++) {
    int y = G[x][i];
    if (dp[y]) dfs2(y);
  }
  printf("%d\n", x);
  for (register unsigned i = 0; i < G[x].size(); i++) {
    int y = G[x][i];
    if (!dp[y]) dfs2(y);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &f[i]), G[f[i]].push_back(i);
  for (int i = 1; i <= n; i++)
    if (!f[i]) {
      rt = i;
      break;
    }
  dfs(rt);
  if (!dp[rt]) {
    puts("YES");
    dfs2(rt);
  } else
    puts("NO");
  return 0;
}
