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;
int n, uu, hea[200005], cnt, deg[200005], fa[200005];
bool vis[200005];
vector<int> shu;
vector<int> ans;
struct Edge {
  int too, nxt;
} edge[400005];
void add_edge(int fro, int too) {
  edge[++cnt].nxt = hea[fro];
  edge[cnt].too = too;
  hea[fro] = cnt;
}
void dfs(int x, int f) {
  shu.push_back(x);
  fa[x] = f;
  for (int i = hea[x]; i; i = edge[i].nxt) {
    int t = edge[i].too;
    if (t != f) dfs(t, x);
  }
}
void shanchu(int x) {
  vis[x] = true;
  ans.push_back(x);
  for (int i = hea[x]; i; i = edge[i].nxt) {
    int t = edge[i].too;
    deg[t]--;
    if (t != fa[x] && !vis[t]) {
      if (deg[t] % 2 == 0) shanchu(t);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &uu);
    if (!uu) continue;
    add_edge(uu, i);
    add_edge(i, uu);
    deg[i]++;
    deg[uu]++;
  }
  dfs(1, 0);
  for (int i = shu.size() - 1; i >= 0; i--) {
    int x = shu[i];
    if (deg[x] & 1) continue;
    shanchu(x);
  }
  if (ans.size() != n)
    printf("NO\n");
  else {
    printf("YES\n");
    for (auto i : ans) printf("%d\n", i);
  }
  return 0;
}