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 to[3000010], last[3000010], nxt[3000010], tot = 0;
int ru[3000010];
bool vis[3000010];
void add(int x, int y) {
  to[++tot] = y;
  nxt[tot] = last[x];
  last[x] = tot;
}
int to2[3000010], last2[3000010], nxt2[3000010], tot2 = 0;
void add2(int x, int y) {
  to2[++tot2] = y;
  nxt2[tot2] = last2[x];
  last2[x] = tot2;
  ru[y]++;
}
int siz[3000010];
void dfs(int x, int pre) {
  siz[x] = 1;
  for (int j = last[x]; j; j = nxt[j]) {
    if (to[j] != pre && to[j] != 0) {
      dfs(to[j], x);
      siz[x] += siz[to[j]];
    }
  }
  if (pre == 0) return;
  if (siz[x] % 2 == 0)
    add2(x, pre);
  else
    add2(pre, x);
}
void work(int x) {
  printf("%d\n", x);
  vis[x] = true;
  for (int j = last[x]; j; j = nxt[j]) {
    ru[to[j]]--;
    if (ru[to[j]] == 0 && vis[to[j]] == false) work(to[j]);
  }
}
int main(int argc, char const *argv[]) {
  memset(ru, 0, sizeof(ru));
  memset(last, 0, sizeof(last));
  memset(vis, false, sizeof(vis));
  int n;
  scanf("%d", &n);
  if (n % 2 == 0) {
    printf("NO\n");
    return 0;
  } else
    printf("YES\n");
  int faa;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    add(x, i);
    add(i, x);
    if (x == 0) faa = i;
  }
  dfs(faa, 0);
  for (int i = 1; i <= n; ++i) {
    if (ru[i] == 0 && vis[i] == false) work(i);
  }
  return 0;
}