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 maxn = 2e5 + 7;
bool ok = true;
int head, tail;
int o[maxn], deg[maxn];
vector<int> e[maxn];
void dfs(int u, int pre) {
  if (pre != 0)
    deg[u] = 1, deg[pre]++;
  else
    deg[u] = 0;
  if (pre != 0 && e[u].size() == 1) {
    o[tail++] = u;
    return;
  }
  for (int i = e[u].size() - 1; i >= 0; i--)
    if (e[u][i] != pre) dfs(e[u][i], u);
  if (deg[u] & 1) {
    if (pre == 0) {
      ok = false;
    }
    o[tail++] = u;
  } else {
    o[--head] = u;
    if (pre) deg[pre]--;
  }
}
int main() {
  int n, root;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    if (u == 0)
      root = i;
    else {
      e[u].push_back(i);
      e[i].push_back(u);
    }
  }
  head = n;
  dfs(1, 0);
  if (!ok) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (int i = n - 1; i >= 0; i--) printf("%d\n", o[i]);
  return 0;
}