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.

<image>
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200010;
bool vis[MAX];
stack<int> sta;
vector<int> G[MAX], res;
int n, t, tem, deg[MAX], f[MAX];
void DFS(int x, int y) {
  sta.push(x);
  f[x] = y;
  for (int i = 0; i < G[x].size(); ++i) {
    if (G[x][i] == y) continue;
    DFS(G[x][i], x);
  }
}
void DFS2(int x) {
  res.push_back(x);
  vis[x] = true;
  for (int i = 0; i < G[x].size(); ++i) {
    int nex = G[x][i];
    --deg[nex];
    if (nex == f[x]) continue;
    if (vis[nex]) continue;
    if (deg[nex] % 2 == 0) DFS2(nex);
  }
}
int main() {
  scanf("%d", &n);
  if (n & 1) {
    puts("YES");
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &t);
      if (!t) continue;
      ++deg[t], ++deg[i];
      G[t].push_back(i);
      G[i].push_back(t);
    }
    DFS(1, 0);
    while (!sta.empty()) {
      tem = sta.top();
      sta.pop();
      if (deg[tem] % 2 == 0) DFS2(tem);
    }
    for (int i = 0; i < res.size(); ++i) printf("%d\n", res[i]);
  } else
    puts("NO");
  return 0;
}
