CodeForces - 963B Destruction of a Tree （思维题）

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.

题目大意：给你一棵树，只能删除度数为偶数的节点，节点删除后，与它相连的边也会删除。问你能否把所有点删除。

解题思路：只要你能想到，如果一棵树，有偶数条边，那么他一定能被删除完！或者说，一棵树有奇数个节点，那么他肯定能被删除完！因为，如果边为奇数，每次删除偶数条边，最后肯定剩奇数个边啊！如果边为偶数，每次删除偶数条边，最后肯定能删除完！所以基于这个思想，我们递归的删除点即可。从根节点开始，如果某一棵子树他的节点个数为偶数（加上当前节点就为奇数了），那么就深搜这颗子树，递归删除。

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

vector<int> ch[200005];
int sz[200005];

void getsize(int u, int pre)
{
	sz[u] = 1;
	for (int i = 0; i < ch[u].size(); ++i)
	{
		if (ch[u][i] != pre)
		{
			getsize(ch[u][i], u);
			sz[u] += sz[ch[u][i]];
		}
	}
}

void dfs(int u, int pre)
{
	for (int i = 0; i < ch[u].size(); i++)
	{
		if (ch[u][i] != pre)
		{
			if (sz[ch[u][i]] % 2 == 0)
			{
				dfs(ch[u][i], u);
			}
		}
	}

	printf("%d\n", u);

	for (int i = 0; i < ch[u].size(); i++)
	{
		if (ch[u][i] != pre)
		{
			if (sz[ch[u][i]] % 2 == 1)
			{
				dfs(ch[u][i], u);
			}
		}
	}
}

int main()
{

	int N;
	scanf("%d", &N);
	int temp;
	int root;
	for (int i = 1; i <= N; i++)
	{
		scanf("%d", &temp);
		if (temp != 0)
		{
			ch[i].push_back(temp);
			ch[temp].push_back(i);
		}
		else
		{
			root = i;
		}
	}

	if (N % 2 == 0)
	{
		printf("NO\n");
	}
	else
	{
		getsize(root,-1);
		printf("YES\n");
		dfs(root, -1);
	}

	return 0;
}