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;
const int MAXN = (int)2e5;
vector<int> g[MAXN + 1];
bool dp[MAXN + 1][2];
bool vis[MAXN + 1][2];
bool dfs(int nod, bool t, int p) {
  if (vis[nod][t]) return dp[nod][t];
  int cnt = 0;
  for (auto it : g[nod]) {
    if (it != p && 2 * dfs(it, 0, nod) + dfs(it, 1, nod) > 1) cnt++;
  }
  vis[nod][t] = 1;
  if (cnt == 0) {
    return dp[nod][t] = (t == 0);
  }
  if ((cnt + t) % 2 == 1) {
    for (auto it : g[nod]) {
      if (it != p && 2 * dp[it][0] + dp[it][1] == 3) {
        cnt--;
        break;
      }
    }
  }
  if ((cnt + t) % 2 == 0) {
    dp[nod][t] = 1;
  }
  return dp[nod][t];
}
void solve(int nod, bool t, int p) {
  int cnt = 0;
  for (auto it : g[nod]) {
    if (it != p && 2 * dp[it][0] + dp[it][1] == 1) {
      solve(it, 1, nod);
    }
    if (it != p && 2 * dp[it][0] + dp[it][1] > 1) {
      cnt++;
    }
  }
  int id = -1;
  if ((cnt + t) % 2 == 1) {
    for (auto it : g[nod]) {
      if (it != p && 2 * dp[it][0] + dp[it][1] == 3) {
        solve(it, 1, nod);
        id = it;
        cnt--;
        break;
      }
    }
  }
  cout << nod << "\n";
  for (auto it : g[nod]) {
    if (it != p && it != id && 2 * dp[it][0] + dp[it][1] != 1) {
      solve(it, 0, nod);
    }
  }
}
int main() {
  int i, n, x;
  ios::sync_with_stdio(false);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    if (x != 0) {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  if (1 - n & 1) {
    cout << "NO";
    return 0;
  }
  cout << "YES"
       << "\n";
  dfs(1, 0, 0);
  solve(1, 0, 0);
  return 0;
}
