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.

<image>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200055;
vector<int> e[maxn], ans, sing;
int deg[maxn], nx[maxn], pre[maxn];
void Erase(int i) {
  nx[pre[i]] = nx[i];
  pre[nx[i]] = pre[i];
}
int getPos() {
  int x = nx[0];
  while (x != -1) {
    if (deg[x] % 2 == 0) return x;
    x = nx[x];
  }
  return -1;
}
vector<int> row[maxn];
int depth[maxn];
bool bad[maxn];
void dfs(int u, int d, int par) {
  row[d].push_back(u);
  depth[u] = d;
  for (int i = 0; i < ((int)e[u].size()); ++i) {
    int v = e[u][i];
    if (v != par) dfs(v, d + 1, u);
  }
}
struct node {
  int x, deg, dep;
  node(int _x = 1, int _deg = 1, int _dep = 1) : x(_x), deg(_deg), dep(_dep) {}
  bool operator<(const node &t) const {
    if (deg % 2 != t.deg % 2) return deg % 2 < t.deg % 2;
    if (dep != t.dep) return depth[x] > depth[t.x];
    return x < t.x;
  }
};
bool mcmp(int a, int b) { return depth[a] > depth[b]; }
int ver[maxn];
int main() {
  int n;
  scanf("%d", &n);
  int rt = -1;
  for (int i = 1; i < (n + 1); ++i) {
    ver[i] = i;
    int x;
    scanf("%d", &x);
    if (x == 0)
      rt = i;
    else {
      e[i].push_back(x);
      deg[x]++;
      e[x].push_back(i);
      deg[i]++;
    }
  }
  dfs(rt, 0, -1);
  set<node> rec;
  for (int i = 1; i < (n + 1); ++i) rec.insert(node(i, deg[i] % 2, depth[i]));
  node t = *rec.begin();
  while ((int)rec.size()) {
    node cur = *rec.begin();
    int x = cur.x;
    if (cur.deg % 2 == 1) break;
    ans.push_back(x);
    rec.erase(rec.begin());
    bad[x] = 1;
    for (int i = 0; i < ((int)e[x].size()); ++i) {
      int v = e[x][i];
      if (bad[v]) continue;
      rec.erase(node(v, deg[v], depth[v]));
      deg[v]--;
      rec.insert(node(v, deg[v], depth[v]));
    }
  }
  if ((int)ans.size() == n) {
    cout << "YES" << endl;
    for (int i = 0; i < ((int)ans.size()); ++i) printf("%d\n", ans[i]);
  } else
    cout << "NO" << endl;
}
