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;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 222222;
vector<int> a[N];
int sz[N];
int par[N];
void dfsSize(int v, int p = -1) {
  for (int x : a[v]) {
    if (x == p) {
      continue;
    }
    dfsSize(x, v);
    sz[v] += sz[x];
  }
  par[v] = p;
  ++sz[v];
}
vector<int> ans;
void dfs(int v) {
  for (int x : a[v]) {
    if (x == par[v]) {
      continue;
    }
    if (sz[x] % 2 == 0) {
      dfs(x);
    }
  }
  ans.push_back(v);
  for (int x : a[v]) {
    if (x == par[v]) {
      continue;
    }
    if (sz[x] % 2) {
      dfs(x);
    }
  }
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n = nxt();
  for (int i = 0; i < n; ++i) {
    int u = nxt() - 1, v = i;
    if (u == -1) {
      continue;
    }
    a[u].push_back(v);
    a[v].push_back(u);
  }
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  dfsSize(0);
  dfs(0);
  printf("YES\n");
  for (int x : ans) {
    printf("%d\n", x + 1);
  }
  return 0;
}