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 long long int maxn = 5 * 100000 + 50, inf = 1e13, N = 3050;
long long int par[maxn];
vector<long long int> v[maxn];
long long int dp[maxn];
vector<long long int> p;
bool f[maxn];
void dfs(long long int x) {
  long long int y = 0;
  p.push_back(x);
  for (long long int i = 0; i < v[x].size(); i++) dfs(v[x][i]);
  for (long long int i = 0; i < v[x].size(); i++) {
    if (dp[v[x][i]] == 1) y++;
  }
  if (y % 2 == 1) {
    dp[x] = 2;
    f[x] = true;
  } else
    dp[x] = 1;
}
int main() {
  long long int n;
  cin >> n;
  long long int rishe = -1;
  for (long long int i = 1; i <= n; i++) {
    long long int c;
    cin >> c;
    if (c == 0) {
      rishe = i;
      continue;
    }
    v[c].push_back(i);
    par[i] = c;
  }
  for (long long int i = 1; i <= n; i++)
    if (v[i].size() == 0) dp[i] = 1;
  dfs(rishe);
  if (dp[rishe] == 2) {
    cout << "NO";
  } else {
    cout << "YES" << endl;
    for (long long int i = p.size() - 1; i > -1; i--) {
      if (f[p[i]] == true) cout << p[i] << endl;
    }
    for (long long int i = 0; i < p.size(); i++) {
      if (f[p[i]] == true) continue;
      cout << p[i] << endl;
    }
  }
  return 0;
}