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 = 200005;
struct Gragh {
  int cnt, y[N * 2], nxt[N * 2], fst[N];
  void clear() {
    cnt = 0;
    memset(fst, 0, sizeof fst);
  }
  void add(int a, int b) { y[++cnt] = b, nxt[cnt] = fst[a], fst[a] = cnt; }
} g;
int n, dp[N];
vector<int> sted[N];
bool spec_cmp(int a, int b) { return dp[a] < dp[b]; }
void solve(int x, int pre) {
  int cnt = 1;
  sted[x].clear();
  for (int i = g.fst[x]; i; i = g.nxt[i])
    if (g.y[i] != pre) {
      solve(g.y[i], x);
      cnt++;
      sted[x].push_back(g.y[i]);
    }
  sort(sted[x].begin(), sted[x].end(), spec_cmp);
  if (sted[x].size() == 0) {
    dp[x] = 1;
    return;
  }
  for (int i = 0; i < sted[x].size(); i++) {
    if (dp[sted[x][i]]) break;
    cnt--;
  }
  dp[x] = cnt & 1;
}
void dfs2(int x) {
  int i = 0;
  while (i < sted[x].size() && dp[sted[x][i]] == 0) dfs2(sted[x][i]), i++;
  printf("%d\n", x);
  while (i < sted[x].size()) dfs2(sted[x][i]), i++;
}
int main() {
  g.clear();
  scanf("%d", &n);
  for (int i = 1, p; i <= n; i++) {
    scanf("%d", &p);
    if (p != 0) g.add(i, p), g.add(p, i);
  }
  solve(1, 0);
  if (dp[1] == 1) {
    puts("YES");
    dfs2(1);
  } else
    puts("NO");
  return 0;
}