Bear and Friendship Condition-HZUN寒假集训

Bear and Friendship Condition

time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output

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 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.

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).

Examples

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

 

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define LL long long  
int pre[151000];  
int Find(int x){  
    return x==pre[x]?x:pre[x]=Find(pre[x]);
}  
void mix(int x,int y){  
    int xx=Find(x),yy=Find(y);  
    if(xx!=yy){  
        pre[xx]=yy;  
    }  
}  
LL t[151000];  
int main(){  
    int n,m;  
    scanf("%d%d",&n,&m);  
    for(int i=1;i<=n;i++){  
        pre[i]=i;  
    }  
    for(int i=0;i<m;i++){  
        int x,y;  
        scanf("%d%d",&x,&y);  
        mix(x,y);  
    }  
    for(int i=1;i<=n;i++){  
        t[Find(i)]++;  
    }  
    LL sum=0;  
    for(int i=1;i<=n;i++){  
        if(t[i]!=0&&t[i]!=1){  
            sum+=(t[i]*(t[i]-1)/2);  
        }  
    }  
    if(sum!=m){  
        printf("NO\n");  
    }  
    else{  
        printf("YES\n");  
    }  
    return 0;  
}