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, root, p[200001];
vector<int> ans, ch[200001], before[200001], after[200001];
bool vis[200001], par[200001];
void dfs(int u) {
  if (ch[u].size() == 0) {
    par[u] = false;
    return;
  }
  for (int c : ch[u]) {
    dfs(c);
    if (par[c])
      before[u].push_back(c);
    else
      after[u].push_back(c);
  }
  par[u] = after[u].size() % 2;
}
void get_ans(int u) {
  for (int c : before[u]) get_ans(c);
  ans.push_back(u);
  for (int c : after[u]) get_ans(c);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    ch[i] = vector<int>();
    before[i] = vector<int>();
    after[i] = vector<int>();
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
    if (p[i] >= 0)
      ch[p[i]].push_back(i);
    else
      root = i;
  }
  dfs(root);
  if (par[root]) {
    printf("NO\n");
    return 0;
  }
  ans = vector<int>();
  get_ans(root);
  printf("YES\n%d", ans[0] + 1);
  for (int i = 1; i < n; i++) printf(" %d", ans[i] + 1);
  printf("\n");
}