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.

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 ai and bi (1?≤?ai,?bi?≤?n,?ai?≠?bi). Members ai and bi are friends with each other. No pair of members will appear more than once in the input.

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).

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：给我们一种朋友关系，必须是a和b是朋友，b和c是朋友，c和a是朋友才满足要求

解法：对于每一个联通块里面的点，其必须与其他在联通块的点都相连，就是联通块的点个数-1,否则不符合要求

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int dr[200000];
 4 vector<int>q[200000];
 5 int flag=0;
 6 int n,m;
 7 int vis[200000];
 8 queue<int>p;
 9 void dfs(int v)
10 {
11     if(vis[v]==1)
12     {
13         return;
14     }
15     vis[v]=1;
16   //  cout<<v<<endl;
17     p.push(v);
18     for(int i=0;i<q[v].size();i++)
19     {
20         int pos=q[v][i];
21         if(dr[pos]!=dr[v])
22         {
23            // flag=1;
24         }
25         if(vis[pos]==0)
26         {
27             //vis[pos]=1;
28             dfs(pos);
29         }
30     }
31 }
32 int main()
33 {
34     cin>>n>>m;
35     for(int i=1;i<=m;i++)
36     {
37         int s,e;
38         cin>>s>>e;
39         q[s].push_back(e);
40         q[e].push_back(s);
41         dr[s]++;
42         dr[e]++;
43     }
44     for(int i=1;i<=n;i++)
45     {
46         if(vis[i]==0)
47         {
48            // cout<<endl;
49             dfs(i);
50             int cnt=p.size();
51             while(!p.empty())
52             {
53                 int x=p.front();
54                 if(dr[x]!=cnt-1)
55                 {
56                     flag=1;
57                 }
58                 p.pop();
59             }
60         }
61     }
62     if(flag==1)
63     {
64         cout<<"NO"<<endl;
65     }
66     else
67     {
68         cout<<"YES"<<endl;
69     }
70     return 0;
71 }