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 = 500000 + 5;
int head[maxn], to[maxn << 1], nex[maxn << 1], cnt, degree[maxn], parent[maxn],
    vis[maxn];
stack<int> dfs;
queue<int> answer;
inline void add_edge(int u, int v) {
  nex[++cnt] = head[u];
  head[u] = cnt;
  to[cnt] = v;
}
void dfs1(int u, int fa) {
  dfs.push(u);
  parent[u] = fa;
  for (int v = head[u]; v; v = nex[v])
    if (to[v] != fa) dfs1(to[v], u);
}
void dfs2(int u) {
  answer.push(u);
  vis[u] = 1;
  for (int v = head[u]; v; v = nex[v]) {
    --degree[to[v]];
    if (to[v] == parent[u] || vis[to[v]]) continue;
    if (degree[to[v]] % 2 == 0) dfs2(to[v]);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int a;
    scanf("%d", &a);
    if (a == 0) continue;
    add_edge(i, a);
    add_edge(a, i);
    ++degree[a], ++degree[i];
  }
  dfs1(1, 0);
  while (dfs.size()) {
    int u = dfs.top();
    dfs.pop();
    if (degree[u] % 2 == 0) dfs2(u);
  }
  if (answer.size() == n) {
    printf("YES\n");
    while (!answer.empty()) {
      printf("%d\n", answer.front());
      answer.pop();
    }
  } else
    printf("NO");
  return 0;
}