Codeforces Round #475 (Div. 2) D. Destruction of a Tree

D. Destruction of a Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

5
0 1 2 1 2

output

Copy

YES
1
2
3
5
4

input

Copy

4
0 1 2 3

output

Copy

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：给你一颗有n个节点，n-1条边的树。在这棵树中度为偶数的节点可以删除，问是否可以把这棵树中所有节点全部删除。思路：首先我们可以确定的是，如果这棵树的节点个数为偶数的话，那么肯定就不能删除干净了。（因为节点为偶数个数，那边肯定就是基数了，只有度为偶数的节点才能删除，那么到时候必定会存在度为奇数的节点了）另外删点一定要从最靠近叶子节点的偶数度节点开始，如果找到了最靠近叶子节点的偶数度节点，这个时候以它为根节点的子树中的节点肯定都是度为奇数的，那么这个最靠近叶子节点的偶数度节点如果晚于它的父节点被删除的话（假设它的父节点为偶数节点），那么此时度减1就变成奇数了，那么这个节点包括它的子树中的节点就没有可能被删除了。所以我们从从下至上来删点，每次删完一个点之后它的子节点肯定会受到影响，我们在向下查询是否新产生了偶数度节点可以删除。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;
const int MAXN=200005;
int a[MAXN],deg[MAXN],n,pre[MAXN],vis[MAXN];
vector<int> ans;
vector<int> edge[MAXN];
stack<int> sta;
void dfs(int p,int x)
{
    pre[x]=p;
    sta.push(x);
    for(int i=0;i<edge[x].size();i++)
    {
        int y=edge[x][i];
        if(y==p)
            continue;
        dfs(x,y);
    }
}
void dfs2(int x)
{
    vis[x]=1;
    if(deg[x]%2==0)
    ans.push_back(x);
    for(int i=0;i<edge[x].size();i++)
    {
        int y=edge[x][i];
        deg[y]--;
        if(pre[x]==y||vis[y])
            continue;
        dfs2(y);
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            //建树并确定节点度数
            if(a[i])
            {
                edge[a[i]].push_back(i);
                edge[i].push_back(a[i]);
                deg[a[i]]++;
                deg[i]++;
            }
        }
        //把节点放入栈中，方便我们从叶子节点开始找起
        dfs(0,1);
        int len=sta.size();
        for(int i=0;i<len;i++)
        {
            int k=sta.top();
            sta.pop();
            if(deg[k]%2==0)
                dfs2(k);
        }
        if(ans.size()==n)//如果被删除的节点的个数是n，代表全部被删除了
        {
            printf("YES\n");
            for(int i=0;i<ans.size();i++)
                printf("%d\n",ans[i]);
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}