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;
const int MAXN = 5e5 + 5;
struct edge {
  int to, next;
} e[MAXN << 1], dag[MAXN << 1];
int head[MAXN], cnt = 0, n, headdag[MAXN], cntdag = 0, in[MAXN];
inline void add(int u, int v) {
  e[++cnt] = (edge){v, head[u]}, head[u] = cnt;
  e[++cnt] = (edge){u, head[v]}, head[v] = cnt;
}
inline void adddag(int u, int v) {
  dag[++cntdag] = (edge){v, headdag[u]}, headdag[u] = cntdag, in[v]++;
}
int size[MAXN], dep[MAXN];
void dfs(int u, int fa) {
  size[u] = 1;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(v, u);
    size[u] += size[v];
  }
}
void getin(int u, int fa) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    if (size[v] & 1)
      adddag(u, v);
    else
      adddag(v, u);
    getin(v, u);
  }
}
queue<int> q;
void getans() {
  puts("YES");
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (q.size()) {
    int u = q.front();
    q.pop();
    printf("%d\n", u);
    for (int i = headdag[u]; i; i = dag[i].next) {
      int v = dag[i].to;
      in[v]--;
      if (!in[v]) q.push(v);
    }
  }
}
int main() {
  scanf("%d", &n);
  if (!(n & 1)) {
    puts("NO");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    if (tmp) add(tmp, i);
  }
  dfs(1, 1);
  getin(1, 1);
  getans();
  return 0;
}