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;
const int maxn = 200050;
struct Edge {
  int v, nxt;
} e[maxn * 2];
int h[maxn], tot;
void addEdge(int x, int y) {
  e[++tot] = (Edge){y, h[x]};
  h[x] = tot;
}
int n, deg[maxn];
bool vis[maxn];
int ans[maxn], c;
void del(int x, int x_fa) {
  if (vis[x]) return;
  vis[x] = true;
  ans[c++] = x;
  for (int i = h[x]; i; i = e[i].nxt) {
    deg[e[i].v]--;
  }
  for (int i = h[x]; i; i = e[i].nxt) {
    if (e[i].v != x_fa && !vis[e[i].v]) {
      del(e[i].v, x);
    }
  }
}
void dfs(int x, int x_fa) {
  for (int i = h[x]; i; i = e[i].nxt) {
    if (e[i].v == x_fa) continue;
    dfs(e[i].v, x);
  }
  if (deg[x] % 2 == 0) del(x, x_fa);
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (x) {
      addEdge(x, i);
      addEdge(i, x);
      deg[x]++;
      deg[i]++;
    } else
      root = i;
  }
  dfs(root, 0);
  if (c < n) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (int i = 0; i < n; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}