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>
const int MAXN = 2e5 + 5;
int n, root, p[MAXN], mem[MAXN][2], vis[MAXN];
std::vector<int> adj[MAXN], ch[MAXN][2][2];
std::queue<int> order;
int solve(int u, int x) {
  if (mem[u][x] != -1) return mem[u][x];
  if (!vis[u]) {
    for (int v : adj[u]) {
      int a = solve(v, 0), b = solve(v, 1);
      ch[u][a][b].push_back(v);
    }
    vis[u] = 1;
  }
  if (!ch[u][0][0].empty())
    mem[u][x] = 0;
  else if ((ch[u][1][1].size() + ch[u][1][0].size() + x) % 2 == 0)
    mem[u][x] = 1;
  else if (!ch[u][1][1].empty())
    mem[u][x] = 1;
  else
    mem[u][x] = 0;
  return mem[u][x];
}
void reconstruct(int u, int x) {
  for (int v : ch[u][0][1]) reconstruct(v, 1);
  if ((ch[u][1][1].size() + ch[u][1][0].size() + x) % 2 == 0) {
    order.push(u);
    for (int v : ch[u][1][1]) reconstruct(v, 0);
    for (int v : ch[u][1][0]) reconstruct(v, 0);
  } else {
    reconstruct(ch[u][1][1][0], 1);
    order.push(u);
    for (size_t i = 1; i < ch[u][1][1].size(); ++i) {
      reconstruct(ch[u][1][1][i], 0);
    }
    for (int v : ch[u][1][0]) reconstruct(v, 0);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int par;
    scanf("%d", &par);
    if (par == 0) root = i;
    adj[par].push_back(i);
  }
  memset(mem, -1, sizeof mem);
  if (solve(root, 0)) {
    printf("YES\n");
    reconstruct(root, 0);
    for (; !order.empty(); order.pop()) printf("%d ", order.front());
    printf("\n");
  } else {
    printf("NO\n");
  }
}