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.

<image>
#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
vector<int> vv[201000];
int f[201000];
int judge(int u, int pre) {
  int ff = 1;
  for (int i = 0; i < vv[u].size(); ++i) {
    int v = vv[u][i];
    if (v == pre) continue;
    judge(v, u);
    ff ^= f[v];
  }
  f[u] = ff;
  return f[u];
}
void Print(int u, int pre, int t) {
  for (int i = 0; i < vv[u].size(); ++i) {
    int v = vv[u][i];
    if (v == pre) continue;
    if (!f[v]) Print(v, u, 0);
  }
  printf("%d\n", u);
  for (int i = 0; i < vv[u].size(); ++i) {
    int v = vv[u][i];
    if (v == pre) continue;
    if (f[v]) Print(v, u, -1);
  }
}
int main() {
  int n, p;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p);
    if (p) vv[i].push_back(p), vv[p].push_back(i);
  }
  if (judge(1, -1)) {
    printf("YES\n");
    Print(1, -1, 0);
  } else {
    printf("NO\n");
  }
}
