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> edge[200000 + 5];
vector<int> sol;
bool dp[200000 + 5];
bool dfs(int idx, int p = -1) {
  vector<int> D, ND;
  for (auto x : edge[idx])
    if (x != p) {
      if (dfs(x, idx))
        ND.push_back(x);
      else
        D.push_back(x);
    }
  if (ND.empty() && D.empty()) return dp[idx] = true;
  if (ND.size() & 1) return dp[idx] = false;
  return dp[idx] = true;
}
void backT(int idx, int p = -1) {
  for (auto x : edge[idx])
    if (x != p) {
      if (!dp[x]) {
        backT(x, idx);
      }
    }
  sol.push_back(idx);
  for (auto x : edge[idx])
    if (x != p) {
      if (dp[x]) {
        backT(x, idx);
      }
    }
}
int main() {
  int i, n, p;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &p);
    if (p) {
      edge[p].push_back(i);
      edge[i].push_back(p);
    }
  }
  if (dfs(1)) {
    puts("YES");
    backT(1);
    for (auto x : sol) printf("%d\n", x);
  } else
    puts("NO");
  return 0;
}