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;
vector<int> ans;
vector<int> edge[(int)2E5 + 10];
stack<int> sta;
int parent[(int)2E5 + 10];
int flag[(int)2E5 + 10];
int dge[(int)2E5 + 10];
void dfs(int p, int i) {
  parent[i] = p;
  sta.push(i);
  int len = edge[i].size();
  for (int j = 0; j < len; j++) {
    int r = edge[i][j];
    if (r == p) continue;
    dfs(i, r);
  }
}
void dfs2(int i) {
  if (dge[i] % 2 == 0) {
    ans.push_back(i);
    flag[i] = 1;
    for (int j = 0; j < edge[i].size(); j++) {
      int r = edge[i][j];
      dge[r]--;
      if (r == parent[i] || flag[r]) continue;
      dfs2(r);
    }
  } else {
    for (int j = 0; j < edge[i].size(); j++) {
      int r = edge[i][j];
      if (r == parent[i] || flag[r]) continue;
      dfs2(r);
    }
  }
}
int main() {
  int n, x, heap;
  while (~scanf("%d", &n)) {
    memset(flag, 0, sizeof(flag));
    memset(parent, 0, sizeof(parent));
    memset(dge, 0, sizeof(dge));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &x);
      if (x) {
        edge[i].push_back(x);
        edge[x].push_back(i);
        dge[i]++;
        dge[x]++;
      } else
        heap = i;
    }
    dfs(0, heap);
    int len = sta.size();
    for (int i = 1; i <= len; i++) {
      int point = sta.top();
      sta.pop();
      if (dge[point] % 2 == 0) {
        dfs2(point);
      }
    }
    if (ans.size() == n) {
      printf("YES\n");
      for (int i = 0; i < n; i++) printf("%d\n", ans[i]);
    } else
      printf("NO\n");
    vector<int>().swap(ans);
    for (int i = 0; i <= n; i++) vector<int>().swap(edge[i]);
    while (!sta.empty()) sta.pop();
  }
}