Background 
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs. 
Problem

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

Output

The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

Sample Input

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

Hint

Huge input,scanf is recommended.

程式碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int f[2010],s[2010];

void init(int n)
{
	for(int i=0;i<=n;i++)
	f[i]=i;
	memset(s,0,sizeof(s));
}
int getf(int u)
{
	if(f[u]==u)
	return u;
	return f[u]=getf(f[u]);
}
bool merge(int u,int v)
{
	int t1=getf(u);
	int t2=getf(v);
	if(t1==t2)
	return false;
	f[t2]=t1;
	return true;
}

int main(void)
{
	int t,k=0;
	ios::sync_with_stdio(false);
	cin>>t;
	while(t--)
	{
		int n, m;
		cin>>n>>m;
		init(n);
		int u,v,t1,t2;
		int flag=0;
		for(int i=1;i<=m;i++)
		{
			cin>>u>>v;
			if(flag)
			continue;
			t1=getf(u);
			t2=getf(v);
			if(t1==t2)
			{
				flag=1;
				continue;
			}
			//以下操作把同類蟲子連線起來 
			if(s[u]==0)
			s[u]=v;
			else
			merge(s[u],v);//若u已經有物件則把u原來的物件和現在的物件歸為同一類 
			//不用擔心u的舊物件和新物件是否是同一性別，因為前面的t1==t2已經確定了u和u現在的物件不可能是同一性別 
			if(s[v]==0)
			s[v]=u;
			else
			merge(s[v],u);
		}
		cout<<"Scenario #"<<++k<<":"<<endl;
		if(flag)
		cout<<"Suspicious bugs found!"<<endl<<endl;
		else
		cout<<"No suspicious bugs found!"<<endl<<endl;
	}
	return 0;
}