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 = int(2e5) + 5;
int dp[maxn][2], memoi[maxn][2];
vector<int> graph[maxn];
void dfs(int node) {
  for (auto it : graph[node]) dfs(it);
  for (int p = 0; p < 2; p++) {
    int cnt0 = 0, cnt1 = 0, cnt01 = 0;
    for (auto it : graph[node]) {
      if (dp[it][0] && dp[it][1])
        cnt01++;
      else if (dp[it][0])
        cnt0++;
      else if (dp[it][1])
        cnt1++;
      else {
        dp[node][p] = 0;
        continue;
      }
    }
    for (int i = 0; i <= cnt01; i++) {
      int v = int(graph[node].size()) + 1 - (!p) - cnt1 - i;
      if (v % 2 == 0) dp[node][p] = 1, memoi[node][p] = i;
    }
  }
}
void dfs2(int node, int p) {
  vector<int> cnt0, cnt1, cnt01;
  for (auto it : graph[node]) {
    if (dp[it][0] && dp[it][1])
      cnt01.push_back(it);
    else if (dp[it][0])
      cnt0.push_back(it);
    else if (dp[it][1])
      cnt1.push_back(it);
  }
  for (auto it : cnt1) dfs2(it, 1);
  for (int i = 0; i < memoi[node][p]; i++) dfs2(cnt01[i], 1);
  printf("%d\n", node + 1);
  for (auto it : cnt0) dfs2(it, 0);
}
int main(void) {
  int n, p, root = -1;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p);
    p--;
    if (p == -1)
      root = i;
    else
      graph[p].push_back(i);
  }
  dfs(root);
  if (dp[root][0]) {
    printf("YES\n");
    dfs2(root, 0);
  } else
    printf("NO\n");
}