Codeforces791 B. Bear and Friendship Condition

 

B. 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

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.

 

 

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题目的意思是给出n个点m条边，问每个点是否都落在一个完全图中

思路：先并查集处理出所有集合，在数学算出每个集合成为完全图所需的边总和和m比较

注意：题目会爆int

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
int pre[200005];
LL cnt[200005];
void init()
{
    for(int i=0; i<200004; i++)
        pre[i]=i;
}

int fin(int x)
{
    return pre[x]==x?x:pre[x]=fin(pre[x]);
}

int main()
{
    int n,m,x,y;
    scanf("%d%d",&n,&m);
    init();
    for(int i=0; i<m; i++)
    {
        scanf("%d%d",&x,&y);
        int a=fin(x);
        int b=fin(y);
        if(a!=b)
        {
            pre[a]=b;
        }
    }
    memset(cnt,0,sizeof cnt);
    for(int i=1;i<=n;i++)
    {
        int x=fin(i);
        cnt[x]++;
    }
    LL ans=0;
     for(int i=1;i<=n;i++)
     {
         ans+=(cnt[i]*(cnt[i]-1)/2);
     }
     printf("%s\n",ans==1LL*m?"YES":"NO");

    return 0;
}