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 N = 5e5 + 50;
int E, V[N], Next[N], Head[N];
bool vis[N];
void add(int a, int b) {
  V[E] = b;
  Next[E] = Head[a];
  Head[a] = E++;
}
bool dfs(int u, int fa) {
  bool ans = 0;
  for (int i = Head[u]; i != -1; i = Next[i]) {
    int v = V[i];
    if (v == fa) continue;
    vis[v] = dfs(v, u);
    if (!vis[v]) ans = !ans;
  }
  return ans;
}
void print(int u, int fa) {
  for (int i = Head[u]; i != -1; i = Next[i]) {
    int v = V[i];
    if (v == fa) continue;
    if (vis[v]) print(v, u);
  }
  printf("%d\n", u);
  for (int i = Head[u]; i != -1; i = Next[i]) {
    int v = V[i];
    if (v == fa) continue;
    if (!vis[v]) print(v, u);
  }
}
int main() {
  int n, rt;
  memset(Head, -1, sizeof(Head));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p == 0) {
      rt = i;
      continue;
    }
    add(i, p), add(p, i);
  }
  if (!dfs(rt, -1)) {
    puts("YES");
    print(rt, -1);
  } else
    puts("NO");
}