Tinkoff Internship Warmup Round 2018 and 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.

题解：
以其中一个点为根，从叶子节点开始遍历。若此节点度数为奇数，
则不操作，若是偶数，则一定要删除，否则，若删除他的父亲节点
则此点变为奇数就不能删掉了。
代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=200005;
int u[maxn],par[maxn];
bool vis[maxn];
vector<int>rode[maxn];
stack<int>P;//存储遍历顺序
queue<int>Q;
void dfs(int v,int pre)
{
    P.push(v);par[v]=pre;
    for(int i=0;i<rode[v].size();i++)
    {
        if(rode[v][i]==pre)continue;
        dfs(rode[v][i],v);
    }
}
void dfs2(int x)
{
    Q.push(x);vis[x]=1;
    for(int i=0;i<rode[x].size();i++)
    {
        int to=rode[x][i];
        u[to]--;
        if(to==par[x]||vis[to])continue;
        if(u[to]%2==0)
            dfs2(to);
    }
}
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int x;scanf("%d",&x);
        if(x)
        {
            rode[i].push_back(x),rode[x].push_back(i);
            u[i]++,u[x]++;
        }
    }
    dfs(1,0);
    while(!P.empty())
    {
        int x=P.top();P.pop();
        //printf("%d\n",x);
        if(u[x]%2==0)
            dfs2(x);
    }
    //printf("%d\n",Q.size());
    if(Q.size()==n)
    {
        printf("YES\n");
        while(!Q.empty())
        {
            printf("%d\n",Q.front());
            Q.pop();
        }
    }
    else printf("NO\n");
    return 0;
}