Codeforces Round #475 (Div. 1) B. Destruction of a Tree

题目：http://codeforces.com/contest/963/problem/B

B. 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.

sol:

注意到每次删点都是删去偶数条边(只能删去度数为偶数的点）, 当n为偶数时无解。

考虑从根开始dfs，然后从底向上处理结点，当处理到结点u时,子树要么为空，要么为所有点度数为奇数的树。若u度数为偶数，以u为根的树可以依次处理，删去。

code1:

#include
#include
#include
#include
#include
#include
#include
#include
#include

code2：

#include

using namespace std;

const int maxn = 2e5+10;
#define pb push_back

vector e[maxn];
vector ord;

int rt,n;
int d[maxn];
int p[maxn];

void dfs(int u){
    for(int v:e[u]){
        dfs(v);
    }
    ord.pb(u);
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&p[i]);
        d[p[i]]++;
        if(p[i]){
            e[p[i]].pb(i);
            d[i]++;
        }else{
            rt=i;
        }
    }
    dfs(rt);
    vector vv;
    vector ans;
    for(int i : ord){
        if(d[i]&1){
            if(i==rt){
                printf("NO\n");
                exit(0);
            }
            vv.pb(i);
        }else{
            ans.pb(i);
            --d[p[i]];
        }
    }
    printf("YES\n");
    for(auto x:ans){
        printf("%d\n",x);
    }
    reverse(vv.begin(),vv.end());
    for(auto x:vv){
        printf("%d\n",x);
    }
    return 0;
}