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 = 2e5 + 10;
vector<int> g[maxn];
int dp[maxn];
int n;
struct Edge {
  int u, v, next;
} edge[maxn * 10];
int head[maxn];
int tot = 0;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void addedge(int u, int v) {
  edge[tot] = Edge{u, v, head[u]};
  head[u] = tot++;
}
void dfs1(int u, int pre) {
  dp[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == pre) continue;
    dfs1(v, u);
    dp[u] += dp[v];
  }
}
int in[maxn];
void dfs2(int u, int pre) {
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (dp[v] % 2 == 0) {
      addedge(v, u);
      in[u]++;
    } else {
      addedge(u, v);
      in[v]++;
    }
    dfs2(v, u);
  }
}
void bfs() {
  cout << "YES\n";
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    cout << u << endl;
    for (int i = head[u]; ~i; i = edge[i].next) {
      int v = edge[i].v;
      in[v]--;
      if (!in[v]) {
        q.push(v);
      }
    }
  }
}
int p[maxn];
int main() {
  cin >> n;
  int rt = -1;
  for (int i = 1; i <= n; i++) cin >> p[i];
  if (n % 2 == 0) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0) rt = i;
    g[p[i]].push_back(i);
  }
  dfs1(rt, -1);
  init();
  dfs2(rt, -1);
  bfs();
  return 0;
}