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 = 1e6 + 10, NX = 1e9 + 7;
int n, m, r, t, a, b, A[N], D[N], M[N];
set<int> V[N];
set<pair<int, int> > S;
void DFS(int v, int p) {
  int c = 0;
  for (auto X : V[v]) {
    if (X == p) continue;
    DFS(X, v);
    c += !M[X];
  }
  if (c & 1) M[v] = 1, c++;
  D[v] -= c;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r);
    if (r) V[i].insert(r), V[r].insert(i);
  }
  if (n % 2 == 0) return !printf("NO");
  for (int i = 1; i <= n; i++) D[i] = V[i].size();
  printf("YES\n");
  DFS(1, 0);
  for (int i = 1; i <= n; i++) S.insert({D[i], i});
  while (S.size()) {
    auto X = *S.begin();
    S.erase(S.begin());
    for (auto v : V[X.second]) {
      V[v].erase(X.second);
      S.erase({D[v], v});
      D[v]--;
      S.insert({D[v], v});
    }
    V[X.second].clear();
    printf("%d\n", X.second);
  }
  return (0);
}