#include <bits/stdc++.h>
using namespace std;
int d[200003];
int dd[200003];
bool vis[200003];
int siz[200003];
int f[200003];
int ans[200003];
queue<int> q;
int n, rt, sum;
struct Edge {
  int to, next;
} edge[400003];
int head[200003];
int cnt;
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
void add(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
void getrt(int u, int fa) {
  siz[u] = 1;
  f[u] = 0;
  bool flag = true;
  for (int i = head[u]; ~i; i = edge[i].next) {
    if (edge[i].to != fa && !vis[edge[i].to]) {
      getrt(edge[i].to, u);
      siz[u] += siz[edge[i].to];
      f[u] = max(f[u], siz[edge[i].to]);
      if (siz[edge[i].to] % 2 == 0) {
        dd[u]++;
        flag = false;
      }
    }
  }
  f[u] = max(f[u], sum - siz[u]);
  if (sum - siz[u] != 0 && (sum - siz[u]) % 2 == 0) dd[u]++;
  if (d[u] % 2 == 0 && flag) q.push(u);
}
void solve(int u) {
  vis[u] = true;
  printf("%d\n", u);
  for (int i = head[u]; ~i; i = edge[i].next) {
    dd[edge[i].to]--;
    if (dd[edge[i].to] == 0) q.push(edge[i].to);
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (v) {
      add(v, i);
      add(i, v);
      d[v]++;
      d[i]++;
    }
  }
  if (n % 2 == 0) {
    printf("NO\n");
  } else {
    f[0] = n + 1, sum = n;
    printf("YES\n");
    getrt(1, 0);
    while (!q.empty()) {
      rt = q.front();
      q.pop();
      solve(rt);
    }
  }
}

### Prompt

Create a solution in CPP for the following 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.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int d[200003];
int dd[200003];
bool vis[200003];
int siz[200003];
int f[200003];
int ans[200003];
queue<int> q;
int n, rt, sum;
struct Edge {
  int to, next;
} edge[400003];
int head[200003];
int cnt;
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
void add(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
void getrt(int u, int fa) {
  siz[u] = 1;
  f[u] = 0;
  bool flag = true;
  for (int i = head[u]; ~i; i = edge[i].next) {
    if (edge[i].to != fa && !vis[edge[i].to]) {
      getrt(edge[i].to, u);
      siz[u] += siz[edge[i].to];
      f[u] = max(f[u], siz[edge[i].to]);
      if (siz[edge[i].to] % 2 == 0) {
        dd[u]++;
        flag = false;
      }
    }
  }
  f[u] = max(f[u], sum - siz[u]);
  if (sum - siz[u] != 0 && (sum - siz[u]) % 2 == 0) dd[u]++;
  if (d[u] % 2 == 0 && flag) q.push(u);
}
void solve(int u) {
  vis[u] = true;
  printf("%d\n", u);
  for (int i = head[u]; ~i; i = edge[i].next) {
    dd[edge[i].to]--;
    if (dd[edge[i].to] == 0) q.push(edge[i].to);
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    if (v) {
      add(v, i);
      add(i, v);
      d[v]++;
      d[i]++;
    }
  }
  if (n % 2 == 0) {
    printf("NO\n");
  } else {
    f[0] = n + 1, sum = n;
    printf("YES\n");
    getrt(1, 0);
    while (!q.empty()) {
      rt = q.front();
      q.pop();
      solve(rt);
    }
  }
}

```