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 = 200005;
vector<int> res;
vector<int> dz[N], levels[N];
set<int> visited;
int i, j, n, ojciec[N], w, siz[N], root;
void go(int w) {
  res.push_back(w);
  visited.insert(w);
  for (int i = 0; i < dz[w].size(); i++)
    if (visited.find(dz[w][i]) == visited.end()) go(dz[w][i]);
}
void countLevels(int w, int l) {
  levels[l].push_back(w);
  for (int i = 0; i < dz[w].size(); i++) countLevels(dz[w][i], l + 1);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    dz[i].clear();
    levels[i].clear();
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &ojciec[i]);
    if (ojciec[i] != 0)
      dz[ojciec[i]].push_back(i);
    else
      root = i;
  }
  visited.clear();
  countLevels(root, 0);
  for (i = 1; i <= n; i++) siz[i] = dz[i].size() + 1;
  siz[root]--;
  for (i = n; i >= 1; i--) {
    for (j = 0; j < levels[i].size(); j++) {
      w = levels[i][j];
      if (siz[w] % 2 == 0) {
        go(w);
        siz[ojciec[w]]--;
      }
    }
  }
  if (siz[root] % 2 == 0) {
    go(root);
    printf("YES\n");
    for (i = 0; i < res.size(); i++) printf("%d\n", res[i]);
  } else {
    printf("NO\n");
  }
  return 0;
}