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 double PI = acos(-1.0);
const int N = 2e5 + 7;
vector<int> E[N];
int deg[N], n, des[N], p[N];
vector<int> pos;
int dep[N];
void dfs(int u, int f, int d) {
  dep[u] = d;
  for (auto &v : E[u]) {
    if (v != f) dfs(v, u, d + 1);
  }
}
inline void add(int s, int t) { E[s].push_back(t); }
void bfs() {
  priority_queue<pair<int, int> > Q;
  int i;
  for (i = 1; i <= n; ++i) {
    if (deg[i] & 1) continue;
    Q.push({dep[i], i});
  }
  while (!Q.empty()) {
    int u = Q.top().second;
    Q.pop();
    if (des[u] || ((deg[u] > 0) && (deg[u] & 1))) continue;
    des[u] = 1;
    pos.push_back(u);
    for (auto &v : E[u]) {
      if (des[v] || (deg[v] <= 0)) continue;
      --deg[v];
      --deg[u];
    }
    for (auto &v : E[u]) {
      if (des[v]) continue;
      if (deg[v] % 2 == 0) Q.push({dep[v], v});
    }
  }
}
int main(void) {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
    if (p[i]) {
      add(p[i], i);
      add(i, p[i]);
      ++deg[i];
      ++deg[p[i]];
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!p[i]) dfs(i, 0, 1);
  }
  bfs();
  int ans = pos.size();
  if (ans != n)
    puts("NO");
  else {
    puts("YES");
    for (auto &x : pos) printf("%d\n", x);
  }
  return 0;
}