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.

<image>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
constexpr ll mod = 1000000007;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 + 99);
const vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
template <typename T, typename Y>
ostream& operator<<(ostream& os, const pair<T, Y>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const V<T>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
template <typename... Args>
void debug(Args&... args) {
  for (auto const& x : {args...}) {
    cerr << x << ' ';
  }
  cerr << '\n';
}
VV<int> e;
V<int> ans1(0), ans2(0), deg;
bool dfs(int now, int pre) {
  bool res = false;
  for (auto&& to : e[now]) {
    if (to == pre) continue;
    res |= dfs(to, now);
  }
  if (deg[now] % 2) {
    ans2.push_back(now);
    return true;
  } else {
    ans1.push_back(now);
    for (auto&& to : e[now]) {
      deg[to]--;
    }
    return false;
  }
}
signed main() {
  cin.tie(0);
  cerr.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  int n;
  cin >> n;
  e.assign(n, V<int>(0));
  deg.assign(n, 0);
  for (ll i = 0; i < (ll)(n); ++i) {
    int x;
    cin >> x;
    --x;
    if (x < 0) continue;
    e[i].push_back(x);
    e[x].push_back(i);
    deg[x]++;
    deg[i]++;
  }
  if (!dfs(0, -1)) {
    cout << "YES" << '\n';
    reverse((ans2).begin(), (ans2).end());
    for (ll i = 0; i < (ll)((ll)(ans1).size()); ++i) {
      cout << ans1[i] + 1 << '\n';
    }
    for (ll i = 0; i < (ll)((ll)(ans2).size()); ++i) {
      cout << ans2[i] + 1 << '\n';
    }
  } else {
    cout << "NO" << '\n';
  }
}
