1.26寒假集训（并查集三连←.←）-3

B - Bear and Friendship Condition

 

Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).

There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can't be a friend with themselves.

Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.

For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.

Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.

Input

The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, ) — the number of members and the number of pairs of members that are friends.

The i-th of the next m lines contains two distinct integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). Members a_i and b_i are friends with each other. No pair of members will appear more than once in the input.

Output

If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).

Example

Input

4 3
1 3
3 4
1 4

Output

YES

Input

4 4
3 1
2 3
3 4
1 2

Output

NO

Input

10 4
4 3
5 10
8 9
1 2

Output

YES

Input

3 2
1 2
2 3

Output

NO

Note

The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.

题意：一开始题意搞错了，看了大佬们的写法才反应过来OTZ简单来说就是给出n个人，m组关系。如果A和B是好朋友，B和C是好朋友，那么需要C和A也是好朋友，如果不存在这样的朋友关系，那么就输出NO。因为朋友关系需要是双向的才符合题意，那么一个小的朋友圈子里的每个人连接的朋友数目应该是一样的，而且都为这个小圈子的人数-1（边的条数）。题目来源：https://vjudge.net/contest/209382#problem/B

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;

int fa[155555],t[155555],f[155555];//t存每个点连接的边的条数，f存放每个小集体点的个数

int findfa(int x)//找爸爸
{
	int a=x;
	while(x!=fa[x])
	{
		x=fa[x];
	}
	while(a!=fa[a])
	{
		int z=a;
		a=fa[a];
		fa[z]=x;
	}
	return x;
}

void uni(int a,int b)//合并小朋友
{
	int faa=findfa(a);
	int fab=findfa(b);
	if(faa!=fab)
	{
		fa[faa]=fab;
	}
}

int n,m,temp,temp1,temp2;

int main()
{
	cin>>n>>m;
	for(int i=0;i<=n;i++)
	{
		fa[i]=i;
		t[i]=0;
		f[i]=0;
	}
	for(int i=0;i<m;i++)
	{
		cin>>temp1>>temp2;
		uni(temp1,temp2);
		t[temp1]++,t[temp2]++;
	}
	for(int i=0;i<=n;i++)//计算每个小集体的点的个数
	{
		temp=findfa(i);
		f[temp]++;
	}
	int flag=0;
	for(int i=0;i<=n;i++)
	{
		temp=findfa(i);
		if(f[temp]-1!=t[i])//看点的个数-1是不是和边数一致
		{
			flag=1;
			break;
		}
	}
	if(!flag)
		printf("YES\n");
	else
		printf("NO\n");

	return 0;
}