Codeforces Round #405 B.Bear and Friendship Condition【Dfs+思维】

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 isreasonable if and only if the following condition is satisfied: For every threedistinct members (X,Y,Z), ifX-Y andY-Z then alsoX-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 andm (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 andb_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). Members a_i andb_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).

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.

题目大意：

如果1认识2，2认识3，必须要求有：1认识3.

如果满足上述条件，输出YES,否则输出NO.

思路：

Dfs出一个联通块，记录这个联通块中点的个数：cnt，那么如果这些点的度数都是cnt-1.那么这个连通块就是合法的，否则就是不合法的。

过程随便维护维护就行了。

Ac代码：

#include
#include
#include
#include
using namespace std;
int vis[150500];
int degree[150500];
vectormp[150500];
int flag;
queues;
void Dfs(int u)
{
    vis[u]=1;
    s.push(u);
    for(int i=0;i