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 + 5, INF = 1e9 + 7;
int rd() {
  char ch = getchar();
  int ret = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') ret = ret * 10 + ch - '0', ch = getchar();
  return ret;
}
int n, lat[N];
struct Edge {
  int v, nxt;
} e[N << 1];
int hd[N], cnt;
void addedge(int u, int v) {
  e[++cnt] = (Edge){v, hd[u]};
  hd[u] = cnt;
}
void dfs(int u, int fat = 0) {
  int sum = (fat != 0);
  for (int i = hd[u]; i; i = e[i].nxt)
    if (e[i].v != fat) dfs(e[i].v, u), sum += lat[e[i].v];
  lat[u] = sum & 1;
}
void print(int u, int fat = 0) {
  for (int i = hd[u]; i; i = e[i].nxt)
    if (e[i].v != fat && !lat[e[i].v]) print(e[i].v, u);
  printf("%d\n", u);
  for (int i = hd[u]; i; i = e[i].nxt)
    if (e[i].v != fat && lat[e[i].v]) print(e[i].v, u);
}
int main() {
  int n = rd();
  for (int i = 1; i <= n; i++) {
    int p = rd();
    if (p) addedge(i, p), addedge(p, i);
  }
  dfs(1);
  if (!lat[1]) {
    puts("YES");
    print(1, 0);
    puts("");
  } else
    puts("NO");
  return 0;
}
