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 = 200002;
const int mod = 1e9 + 9;
char O[1 << 14], *S = O, *T = O;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
struct E {
  int y, nex;
} e[N];
int fa[N], fir[N], len, d[N];
void ins(int x, int y) {
  e[++len] = (E){y, fir[x]}, fir[x] = len, ++d[x], ++d[y];
}
int q[N], ql, ans[N], al, dep[N];
bool v[N];
vector<int> p[N];
void dfs(int x) {
  p[dep[x]].push_back(x);
  for (int k = fir[x]; k; k = e[k].nex) {
    int y = e[k].y;
    dep[y] = dep[x] + 1;
    dfs(y);
  }
}
void check(int x) {
  if (d[x] & 1) return;
  ans[++al] = x, v[x] = 1, --d[fa[x]];
  for (int k = fir[x]; k; k = e[k].nex) {
    int y = e[k].y;
    if (!v[y]) {
      --d[y];
      check(y);
    }
  }
}
int main() {
  int n = read(), i, rt;
  for (i = 1; i <= n; ++i) {
    fa[i] = read();
    if (fa[i])
      ins(fa[i], i);
    else
      rt = i;
  }
  dep[rt] = 1;
  dfs(rt);
  for (i = n; i; --i)
    for (int j = 0; j < p[i].size(); ++j) q[++ql] = p[i][j];
  for (i = 1; i <= n; ++i) {
    int x = q[i];
    check(x);
  }
  if (al == n) {
    puts("YES");
    for (i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  } else
    puts("NO");
  return 0;
}