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 = 200007;
const int inf = (1LL << 29);
int head[maxn], to[maxn << 1], next1[maxn << 1], e1;
void addedge(int u, int v) {
  ++e1;
  next1[e1] = head[u];
  head[u] = e1;
  to[e1] = v;
}
int de[maxn], tag;
bool ok[maxn][2];
vector<int> g1[maxn][2], g2[maxn][2];
void print(int u, int x) {
  for (int i = 0; i < g1[u][x].size(); i++) {
    print(g1[u][x][i], 0);
  }
  printf("%d\n", u);
  for (int i = 0; i < g2[u][x].size(); i++) {
    print(g2[u][x][i], 1);
  }
}
void dfs(int u) {
  int sz = 0;
  for (int i = head[u]; i; i = next1[i]) {
    int v = to[i];
    ++sz;
    dfs(v);
  }
  for (int k = 0; k < 2; k++) {
    ok[u][k] = 1;
    int x = 0, y = 0, fi = 0;
    for (int i = head[u]; i; i = next1[i]) {
      int v = to[i];
      if (!ok[v][0] && !ok[v][1])
        ok[u][k] = 0;
      else if (!ok[v][1])
        x++, g1[u][k].push_back(v);
      else
        y++, g2[u][k].push_back(v);
    }
    if (x % 2 != (de[u] - k) % 2 && !(sz - x - y)) {
      ok[u][k] = 0;
    }
    if (x % 2 != (de[u] - k) % 2)
      for (int i = head[u]; i; i = next1[i]) {
        int v = to[i];
        if (!ok[v][0] && !ok[v][1]) {
          if (!fi) g1[u][k].push_back(v), fi = 1;
        }
      }
  }
}
int main() {
  int n;
  cin >> n;
  int rt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x)
      addedge(x, i), ++de[x], ++de[i];
    else
      rt = i;
  }
  dfs(rt);
  if (ok[rt][0])
    printf("YES\n"), print(rt, 0);
  else
    printf("NO");
  return 0;
}
