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;
vector<int> node[200005], ans;
int cnt;
int n, p, rt;
bool vis[200005];
void dfshelper(int x) {
  if (vis[x]) return;
  vis[x] = true;
  ans.push_back(x);
  for (int i = 0; i < node[x].size(); i++) dfshelper(node[x][i]);
}
void dfs(int x) {
  int deg = 0;
  if (x != rt) deg++;
  for (int i = 0; i < node[x].size(); i++) {
    dfs(node[x][i]);
    if (!vis[node[x][i]]) deg++;
  }
  if (deg % 2 == 0) dfshelper(x);
}
int main() {
  scanf("%d", &n);
  memset(vis, 0, sizeof(vis));
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p);
    if (p != 0) cnt++;
    node[p].push_back(i);
    if (p == 0) rt = i;
  }
  if (cnt % 2) {
    printf("NO\n");
    return 0;
  }
  dfs(rt);
  puts("YES");
  for (int i = 0; i < ans.size(); i++) printf("%d\n", ans[i]);
  return 0;
}