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 = 200005;
int u[maxn], par[maxn];
bool vis[maxn];
vector<int> rode[maxn];
stack<int> P;
queue<int> Q;
void dfs(int v, int pre) {
  P.push(v);
  par[v] = pre;
  for (int i = 0; i < rode[v].size(); i++) {
    if (rode[v][i] == pre) continue;
    dfs(rode[v][i], v);
  }
}
void dfs2(int x) {
  Q.push(x);
  vis[x] = 1;
  for (int i = 0; i < rode[x].size(); i++) {
    int to = rode[x][i];
    u[to]--;
    if (to == par[x] || vis[to]) continue;
    if (u[to] % 2 == 0) dfs2(to);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x) {
      rode[i].push_back(x), rode[x].push_back(i);
      u[i]++, u[x]++;
    }
  }
  dfs(1, 0);
  while (!P.empty()) {
    int x = P.top();
    P.pop();
    if (u[x] % 2 == 0) dfs2(x);
  }
  if (Q.size() == n) {
    printf("YES\n");
    while (!Q.empty()) {
      printf("%d\n", Q.front());
      Q.pop();
    }
  } else
    printf("NO\n");
  return 0;
}