## CodeForces 964D Destruction of a Tree （删除树上偶度点 dfs序的巧用）

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.

【题意】

【分析】

【代码】

``````#include<bits/stdc++.h>
using namespace std;

int p[202020];
int du[202020];
int parent[202020];
bool vis[202020];
vector<int>G[202020];
vector<int>vec,ans;
void dfsxu(int u,int fa=-1)
{
vec.push_back(u);
parent[u]=fa;
for(int i=0;i<G[u].size();i++)
if(G[u][i]!=fa)dfsxu(G[u][i],u);
}
void dfsdel(int u)
{
ans.push_back(u);
vis[u]=1;
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
du[v]--;
if(v==parent[u]||vis[v]||du[v]%2)continue;
dfsdel(v);
}
}
int main()
{
int n;
ios::sync_with_stdio(0);
cin>>n;
memset(du,0,sizeof(du));
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
cin>>p[i];
if(p[i])
{
G[i].push_back(p[i]);
G[p[i]].push_back(i);
du[i]++;
du[p[i]]++;
}
}
dfsxu(1);
for(int i=vec.size()-1;i>=0;i--)
{
int v=vec[i];
if(du[v]%2==0)
dfsdel(v);
}

if(ans.size()<n)cout<<"NO"<<endl;
else
{
cout<<"YES\n";
for(int i=0;i<ans.size();i++)
cout<<ans[i]<<endl;
}
return 0;
}
``````