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 + 10;
int root;
int N, Fa[maxn], degree[maxn], vis[maxn] = {};
vector<int> v[maxn << 1], ans;
stack<int> s;
inline long long read() {
  long long s = 0, f = 1;
  char ch;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    s = (s << 3) + (s << 1) + ch - '0';
  return s * f;
}
void dfs(int x, int fa) {
  s.push(x);
  for (int i = 0; i < v[x].size(); ++i) {
    if (v[x][i] == fa) continue;
    dfs(v[x][i], x);
  }
}
void Dfs(int x) {
  ans.push_back(x);
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); ++i) {
    --degree[v[x][i]];
    if (v[x][i] == Fa[x] || vis[v[x][i]]) continue;
    if ((degree[v[x][i]] & 1) == 0) Dfs(v[x][i]);
  }
}
int main() {
  N = read();
  for (int i = 1; i <= N; ++i) {
    Fa[i] = read();
    if (Fa[i])
      v[i].push_back(Fa[i]), v[Fa[i]].push_back(i), ++degree[Fa[i]],
          ++degree[i];
    else
      root = i;
  }
  dfs(root, 0);
  while (!s.empty()) {
    int x = s.top();
    s.pop();
    if (!(degree[x] & 1)) Dfs(x);
  }
  if (ans.size() == N) {
    puts("YES");
    for (int i = 0; i < ans.size(); ++i) printf("%d\n", ans[i]);
  } else
    puts("NO");
}