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 = 2e5 + 10;
int a[N], d[N];
vector<int> G[N];
vector<int> ans;
int vis[N], sz[N];
void init(int fa, int u) {
  sz[u] = 1;
  for (int v : G[u]) {
    if (v != fa) {
      init(u, v);
      sz[u] += sz[v];
    }
  }
}
void dfs(int fa, int u) {
  int flg = 0;
  for (int v : G[u]) {
    if (v != fa) {
      if (sz[v] % 2 == 0) {
        dfs(u, v);
      }
    }
  }
  ans.push_back(u);
  vis[u] = 1;
  for (int v : G[u]) {
    if (v != fa) {
      if (sz[v] % 2) {
        dfs(u, v);
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x) {
      G[i].push_back(x);
      G[x].push_back(i);
      d[x]++;
      d[i]++;
    }
  }
  if (n & 1) {
    init(-1, 1);
    dfs(0, 1);
    puts("YES");
    for (int x : ans) {
      printf("%d\n", x);
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        printf("%d\n", i);
      }
    }
  } else
    puts("NO");
  return 0;
}