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;
constexpr int INF = 1e9 + 5;
constexpr long long INFL = 1e18;
const int maxn = 200005;
int n, root;
vector<int> graf[maxn];
vector<int> skier[maxn];
int par[maxn], degin[maxn], degout[maxn];
void spierpapier() {
  cout << "NO\n";
  exit(0);
}
void dfs(int a) {
  for (int& v : graf[a]) dfs(v);
  if (degout[a] & 1) {
    if (a == root) spierpapier();
    degout[a]++;
    skier[a].push_back(par[a]);
    degin[par[a]]++;
  } else if (a != root) {
    degin[a]++;
    skier[par[a]].push_back(a);
    degout[par[a]]++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    if (p) {
      graf[p].push_back(i);
      par[i] = p;
    } else
      root = i;
  }
  dfs(root);
  cout << "YES\n";
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (degin[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int w = q.front();
    q.pop();
    cout << w << "\n";
    for (int& v : skier[w]) {
      if (--degin[v] == 0) q.push(v);
    }
  }
  return 0;
}