D. Almost Acyclic Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

You are given a directed graph consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.

Can you make this graph acyclic by removing at most one edge from it? A directed graph is called acyclic iff it doesn’t contain any cycle (a non-empty path that starts and ends in the same vertex).

Input

The first line contains two integers n and m (2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.

Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n, u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge from u to v).

Output

If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.

Examples input

3 4
1 2
2 3
3 2
3 1

output

YES

input

5 6
1 2
2 3
3 2
3 1
2 1
4 5

output

NO

Note

In the first example you can remove edge , and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example,  and ) in order to make the graph acyclic.

题意：给一个有向图，判断能否最多删除一条边使得它变成有向无环图。

思路：枚举删去的边很有可能会导致超时，所以我们枚举删去边的弧头节点。删边所导致就是弧头节点入度-1，枚举每个节点的入度-1，如果存在一个节点入度-1后能够成为拓扑序，那么就是YES。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
int read(){  
    int x=0,f=1;char ch=getchar();  
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}  
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();  
    return f*x;  
}

int n,m;
vector<int>g[505];
int ind[505];
queue<int>q;
bool topo(int v,int inds[],int cnt)
{
	int tmp[505];
	for(int i=0;i<505;i++)tmp[i]=inds[i];
	tmp[v]--;
	if(tmp[v])return false;
	while(!q.empty())
		q.pop();
	q.push(v);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		cnt++;
		for(int i=0;i<g[x].size();i++)
		{
			tmp[g[x][i]]--;
			if(!tmp[g[x][i]])q.push(g[x][i]);
		}
	}
	if(cnt==n)return true;
	return false;
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		while(!q.empty())q.pop();
		memset(ind,0,sizeof(ind));
		for(int i=0;i<505;i++)g[i].clear();
		int u,v;
		for(int i=0;i<m;i++)
		{
			u=read();v=read();
			g[u].push_back(v);
			ind[v]++;
		}
		for(int i=1;i<=n;i++)
			if(!ind[i])q.push(i);
		int cnt=0;
		while(!q.empty())
		{
			int x=q.front();
			q.pop();
			cnt++;
			for(int i=0;i<g[x].size();i++)
			{
				ind[g[x][i]]--;
				if(!ind[g[x][i]])q.push(g[x][i]);
			}
		}
		if(cnt==n)
		{
			printf("YES\n");
			continue;
		}
		bool flag=false;
		for(int i=1;i<=n;i++)
		{
			if(ind[i])
			{
				if(topo(i,ind,cnt))
				{
					flag=true;
					printf("YES\n");
					break;
				}
			}
		}
		if(!flag)printf("NO\n");
	}
}