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 MAX = 210000;
const int INF = 1000000000;
const int MOD = 1000000007;
int mod(int a, int b, int m) {
  int ret = 1;
  a %= m;
  while (b) {
    if (b & 1) ret = ret * a % m;
    b >>= 1;
    a = a * a % m;
  }
  return ret;
}
int inv(int a) { return mod(a, MOD - 2, MOD); }
int vabs(int x) { return x > 0 ? x : -x; }
int gcd(int a, int b) { return a % b ? gcd(b, a % b) : b; }
int dir[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
int son[MAX];
bool flag[MAX];
vector<int> e[MAX];
bool used[MAX];
queue<int> q;
int calcSon(int r, int pre) {
  int i;
  son[r] = 1;
  for (i = 0; i < e[r].size(); i++) {
    int v = e[r][i];
    if (v == pre || used[v]) continue;
    son[r] += calcSon(v, r);
  }
  return son[r];
}
void destruction(int r, int pre) {
  int i;
  for (i = 0; i < e[r].size(); i++) {
    int v = e[r][i];
    if (v == pre || son[v] & 1) continue;
    destruction(v, r);
  }
  printf("%d\n", r);
  for (i = 0; i < e[r].size(); i++) {
    int v = e[r][i];
    if (v == pre || !(son[v] & 1)) continue;
    destruction(v, r);
  }
}
int main() {
  int n, i, j;
  cin >> n;
  for (i = 1; i <= n; i++) {
    son[i] = 0;
    e[i].clear();
  }
  for (i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p != 0) {
      e[p].push_back(i);
      e[i].push_back(p);
    }
  }
  if (n % 2 == 0) {
    puts("NO");
    return 0;
  }
  puts("YES");
  calcSon(1, -1);
  destruction(1, -1);
  return 0;
}