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 = 2e5 + 7;
vector<int> vt[MAXN];
int ans[MAXN], sum, root;
bool flag[MAXN];
void dfs2(int u) {
  if (flag[u]) return;
  flag[u] = true;
  ans[sum++] = u;
  for (int i = 0; i < vt[u].size(); i++) dfs2(vt[u][i]);
}
void dfs(int u) {
  int e = 0;
  if (u != root) e++;
  for (int i = 0; i < vt[u].size(); i++) {
    int v = vt[u][i];
    dfs(v);
    if (!flag[v]) e++;
  }
  if (e % 2 == 0) dfs2(u);
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(ans, 0, sizeof(ans));
    memset(flag, false, sizeof(flag));
    sum = 0;
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      if (x == 0)
        root = i;
      else
        vt[x].push_back(i);
    }
    if (n % 2 == 0) {
      printf("NO\n");
      continue;
    }
    dfs(root);
    if (sum == n) {
      printf("YES\n");
      for (int i = 0; i < sum; i++) printf("%d\n", ans[i]);
    } else
      printf("NO\n");
  }
}