In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

Output one line for each test case.
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).

3 2
3 1
2 1
3 3
3 2
2 1
1 3

YES
NO

考查知识点：拓扑排序 

//考查知识点：拓扑排序。。。 
#include<stdio.h>
#include<string.h>
#define inf 110
int line[inf][inf];
int in[inf];
int n,m;
int i,j,k;
int count;
int a,b;
void topo()
{
	for(i=1;i<=n;++i)//n次遍历 
	{
		for(j=1;j<=n;++j)//每次遍历n个数  
		{
			if(!in[j])
			{
				in[j]--;
				count++;
				for(k=1;k<=n;++k)
				{
					if(line[j][k])
					in[k]--;
				} 
			}	
		}
	} 
}
int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		memset(line,0,sizeof(line));
		memset(in,0,sizeof(in));
		count=0;
		for(i=1;i<=m;++i)
		{
			scanf("%d%d",&a,&b);
			if(!line[a][b])
			{
				line[a][b]=1;
				in[b]++;
			}
		}
		topo();
		if(count==n)
		printf("YES\n");
		else
		printf("NO\n");
	}	
	return 0;
}