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 int N = 2e5, M = 4e5;
int head[N + 10], ver[M + 10], nxt[M + 10], tot, sz[N + 10];
void add(int x, int y) {
  ver[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
  ver[++tot] = x;
  nxt[tot] = head[y];
  head[y] = tot;
}
void dfs1(int x, int fa) {
  sz[x] = 1;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == fa) continue;
    dfs1(y, x);
    sz[x] += sz[y];
  }
}
int ans[N + 10], cnt = 0;
void dfs2(int x, int fa) {
  vector<int> odd, even;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == fa) continue;
    if (sz[y] & 1)
      odd.push_back(y);
    else
      even.push_back(y);
  }
  for (int i = 0; i < even.size(); i++) {
    int y = even[i];
    dfs2(y, x);
  }
  ans[++cnt] = x;
  for (int i = 0; i < odd.size(); i++) {
    int y = odd[i];
    dfs2(y, x);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (!x) continue;
    add(i, x);
  }
  dfs1(1, -1);
  if (!(sz[1] & 1))
    printf("NO");
  else {
    printf("YES\n");
    dfs2(1, -1);
    for (int i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
  }
  return 0;
}
