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 = 200010;
struct edge {
  int y, next;
} e[maxn << 1];
int N, root;
int lin[maxn], len, deg[maxn];
bool v[maxn];
int cnt[maxn][2];
int read() {
  int xx = 0, ff = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * ff;
}
inline void inse(int xx, int yy) {
  e[++len].next = lin[xx];
  lin[xx] = len;
  e[len].y = yy;
  deg[xx]++;
}
bool dfs(int x, int f) {
  for (int i = lin[x]; i; i = e[i].next)
    if (e[i].y != f) cnt[x][dfs(e[i].y, x)]++;
  if ((deg[x] - cnt[x][1]) % 2 == 0)
    v[x] = 1;
  else
    v[x] = 0;
  return v[x];
}
void print(int x, int f) {
  for (int i = lin[x]; i; i = e[i].next)
    if (e[i].y != f)
      if (v[e[i].y]) print(e[i].y, x);
  printf("%d\n", x);
  for (int i = lin[x]; i; i = e[i].next)
    if (e[i].y != f)
      if (!v[e[i].y]) print(e[i].y, x);
}
int temp;
int main() {
  N = read();
  for (int i = 1; i <= N; i++) {
    temp = read();
    if (!temp)
      root = i;
    else
      inse(i, temp), inse(temp, i);
  }
  if (dfs(root, 0)) {
    puts("YES");
    print(root, 0);
  } else
    puts("NO");
}