G - Bear and Friendship Condition （并查集）

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 isreasonable if and only
if the following condition is satisfied: For every threedistinct 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).

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.

思路：认识是相互的说明是无向边，由传递性可知这个无向图必须是无向完全图才是YES 
每个联通分支中无向完全图边的条数与顶点数n关系为n*（n-1）/2 。 用并查集求分支，
比以往不同，可以在并查集时增添一个数组记录每个分支元素数目。要注意得用loog long
虽然边数不会越界，但是你计算的边数（完全图的边数）是可能越界的。

#include <iostream>
#include <cstring>
#include <climits>
#include <cmath>
#include <cstdio>
using namespace std;

typedef long long ll;
const int maxn = 151000;

ll pre[maxn];
ll num[maxn]; // 存每个分支数目
ll m;

void init()
{
for(int i = 1; i<=m ;i ++)
pre[i] = i,num[i] = 1;
}

ll find_ (ll x)
{
return pre[x] == x ? x : (pre[x] = find_(pre[x]));
}

void join(int x, int y)
{
ll fx = find_ (x);
ll fy = find_ (y);
if(fx != fy)
{
pre[fy] = fx;
num[fx] += num[fy];
}
}
int main()
{
ll n,tmp;
scanf("%lld%lld",&m,&n);
init();
tmp = n;
while(n --)
{
ll x,y;
scanf("%lld%lld",&x,&y);
join(x,y);
}
ll tol = 0;
for(int i = 1; i <= m ; i++)
{
if(pre[i] == i) tol += (num[i]*(num[i]-1) / 2);
}
if(tmp == tol) printf("YES\n");
else           printf("NO\n");
return 0;
}

         

思路：认识是相互的说明是无向边，由传递性可知这个无向图必须是无向完全图才是YES 
每个联通分支中无向完全图边的条数与顶点数n关系为n*（n-1）/2 。 用并查集求分支，
比以往不同，可以在并查集时增添一个数组记录每个分支元素数目。要注意得用loog long
虽然边数不会越界，但是你计算的边数（完全图的边数）是可能越界的。