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 N = 2e5 + 10;
const int MOD = 1e9 + 9;
int n;
vector<int> v[N];
int pa[N];
int f[N][2];
vector<int> t[N][2];
int q[N];
void BFS() {
  int top = 1, bot = 1;
  q[1] = 1;
  while (top <= bot) {
    int x = q[top++];
    for (auto y : v[x])
      if (y != pa[x]) {
        pa[y] = x;
        q[++bot] = y;
      }
  }
}
void trace(int x, int label) {
  int n1 = v[x].size();
  for (int i = n1 - 1; i >= 0; i--) {
    int y = v[x][i];
    if (y == pa[x]) continue;
    int use = t[x][label][i];
    if (use == 0) trace(y, use);
  }
  cout << x << " ";
  for (int i = n1 - 1; i >= 0; i--) {
    int y = v[x][i];
    if (y == pa[x]) continue;
    int use = t[x][label][i];
    if (use == 1) trace(y, use);
  }
}
int tr[2][2][N];
int cf[2][2], tf[2][2];
void solve() {
  BFS();
  for (int j = n; j >= 1; j--) {
    int cur = q[j];
    memset(cf, 0, sizeof(cf));
    cf[1][0] = cf[0][1] = 1;
    int n1 = v[cur].size();
    for (int i = 0; i < n1; i++) {
      int y = v[cur][i];
      if (y == pa[cur]) continue;
      for (int k0 = 0; k0 <= 1; k0++)
        for (int k1 = 0; k1 <= 1; k1++) tf[k0][k1] = cf[k0][k1];
      memset(cf, 0, sizeof(cf));
      for (int k0 = 0; k0 <= 1; k0++)
        for (int k1 = 0; k1 <= 1; k1++)
          if (tf[k0][k1]) {
            for (int last = 0; last <= 1; last++)
              if (f[y][last]) {
                cf[k0 ^ last][k1] = 1;
                tr[k0 ^ last][k1][i] = last;
              }
          }
    }
    for (int i = 0; i <= 1; i++) {
      f[cur][i] = cf[0][i];
      int st = 0;
      t[cur][i].resize(n1);
      for (int j = n1 - 1; j >= 0; j--) {
        int y = v[cur][j];
        if (y == pa[cur]) continue;
        t[cur][i][j] = tr[st][i][j];
        st ^= tr[st][i][j];
      }
    }
  }
  if (f[1][1]) {
    cout << "YES\n";
    v[1].push_back(0);
    t[1][0].push_back(0);
    t[1][1].push_back(0);
    trace(1, 1);
    return;
  }
  cout << "NO";
}
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << "NO";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x != 0) {
      v[x].push_back(i);
      v[i].push_back(x);
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 1; j++) t[i][j].resize(v[i].size());
  solve();
}