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;
int n;
basic_string<int> e[200005];
bool sa[200005], bez[200005];
void dfs1(int x, int p) {
  bool lose = false;
  int bravo = -1;
  int csa = 0, cbez = 0;
  for (int y : e[x]) {
    if (y != p) {
      dfs1(y, x);
      if (!sa[y] && !bez[y]) {
        lose = true;
      } else if (sa[y] && bez[y]) {
        bravo = y;
      } else if (sa[y]) {
        csa++;
      } else {
        cbez++;
      }
    }
  }
  if (lose) {
    sa[x] = bez[x] = false;
    return;
  }
  if (bravo == -1) {
    if (cbez % 2) {
      sa[x] = true;
      bez[x] = false;
    } else {
      sa[x] = false;
      bez[x] = true;
    }
  } else {
    sa[x] = bez[x] = true;
  }
}
void dfs2(int x, int p, bool gore) {
  basic_string<int> lsa, lbez, lbravo;
  for (int y : e[x]) {
    if (y != p) {
      if (!sa[y] && !bez[y]) {
      } else if (sa[y] && bez[y]) {
        lbravo += y;
      } else if (sa[y]) {
        lsa += y;
      } else {
        lbez += y;
      }
    }
  }
  for (int y : lsa) {
    dfs2(y, x, true);
  }
  while (gore ^ ((lbez.size() + lbravo.size()) % 2 == 1)) {
    dfs2(lbravo.back(), x, true);
    lbravo.pop_back();
  }
  cout << x << '\n';
  for (int y : lbravo) dfs2(y, x, false);
  for (int y : lbez) dfs2(y, x, false);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x) {
      e[i] += x;
      e[x] += i;
    }
  }
  dfs1(1, 1);
  if (!bez[1]) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  dfs2(1, 1, false);
}