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 + 5;
vector<int> G[maxN];
vector<int> ans;
bool vis[maxN];
int n, root;
int nextInt() {
  int res = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    ;
  for (; isdigit(ch); ch = getchar()) {
    res = res * 10 + ch - 48;
  }
  return res;
}
void getAns(int u) {
  if (vis[u]) return;
  vis[u] = true;
  ans.push_back(u);
  for (auto v : G[u]) {
    getAns(v);
  }
}
int dfs(int u) {
  int res = u != root;
  for (auto v : G[u]) {
    res += dfs(v);
  }
  if (res & 1) return 1;
  getAns(u);
  return 0;
}
int main() {
  n = nextInt();
  for (int v = 1; v <= n; v++) {
    int u = nextInt();
    if (u == 0) root = v;
    G[u].push_back(v);
  }
  if (dfs(root))
    puts("NO");
  else {
    puts("YES");
    for (auto p : ans) {
      printf("%d\n", p);
    }
  }
  return 0;
}