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 mod = 1e9 + 7;
int vis[200005], sum[200005], pre[200005], n, start;
vector<int> v[200005];
queue<int> q;
void del(int x) {
  printf("%d\n", x);
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    sum[y]--;
    if (!vis[y] && y != pre[x] && sum[y] % 2 == 0) {
      del(y);
    }
  }
}
void dfs(int p, int x) {
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y != p) dfs(x, y);
  }
  q.push(x);
  pre[x] = p;
}
int main() {
  scanf("%d", &n);
  memset(sum, 0, sizeof(sum));
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x) {
      v[i].push_back(x);
      v[x].push_back(i);
      sum[i]++;
      sum[x]++;
    }
  }
  if (n & 1) {
    printf("YES\n");
    dfs(0, 1);
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      if (sum[x] % 2 == 0) {
        del(x);
      }
    }
  } else {
    printf("NO\n");
  }
  return 0;
}