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 A = 2e5 + 10;
class Gra {
 public:
  int v, next;
} G[A << 2];
int n, head[A], deg[A], dp[A], tot;
vector<int> vec;
void add(int u, int v) {
  G[tot].v = v;
  G[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int u, int pre) {
  int cnt = 0;
  for (int i = head[u]; i != -1; i = G[i].next) {
    int v = G[i].v;
    if (v == pre) continue;
    dfs(v, u);
    if (dp[v] == 1) cnt++;
  }
  if (cnt & 1)
    dp[u] = 0;
  else
    dp[u] = 1;
}
void Get_Ans(int u, int pre) {
  for (int i = head[u]; i != -1; i = G[i].next) {
    int v = G[i].v;
    if (v == pre) continue;
    if (dp[v] == 0) Get_Ans(v, u);
  }
  vec.push_back(u);
  for (int i = head[u]; i != -1; i = G[i].next) {
    int v = G[i].v;
    if (v == pre) continue;
    if (dp[v] == 1) Get_Ans(v, u);
  }
}
int main() {
  tot = 0;
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x == 0) continue;
    add(x, i);
    add(i, x);
    deg[x]++, deg[i]++;
  }
  dfs(1, 1);
  Get_Ans(1, 1);
  if (dp[1]) {
    puts("YES");
    int Siz = vec.size();
    for (int i = 0; i < Siz; i++) {
      printf("%d\n", vec[i]);
    }
  } else
    puts("NO");
  return 0;
}