Tanvir returned home from the contest and got angry after seeing his room dusty. Who likes to see a dusty room after a brain storming programming contest? After checking a bit he found that there is no brush in him room. So, he called Atiq to get a brush. But as usual Atiq refused to come. So, Tanvir decided to go to Atiq's house.

The city they live in is divided by some junctions. The junctions are connected by two way roads. They live in different junctions. And they can go to one junction to other by using the roads only.

Now you are given the map of the city and the distances of the roads. You have to find the minimum distance Tanvir has to travel to reach Atiq's house.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a blank line. The next line contains two integers N (2 ≤ N ≤ 100) and M (0 ≤ M ≤ 1000), means that there are N junctions and M two way roads. Each of the next M lines will contain three integers u v w (1 ≤ u, v ≤ N, w ≤ 1000)

Output

For each case print the case number and the minimum distance Tanvir has to travel to reach Atiq's house. If it's impossible, then print 'Impossible'.

 解題思路：

就是求從1這個起點到n這個終點的最短距離。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;		
int n,m;
int map[200][200];
bool vis[200];
int dis[200];
const int INF=0x3f3f3f3f;
void init()
{
	int i,j;
	for(i=1;i<=150;i++)
	{
		for(j=1;j<=150;j++)
		{
			if(i==j)
			map[i][j]=0;
			else
			map[i][j]=INF;
		}
	}
}
void f(int x)
{
	int i,j;
	memset(vis,false,sizeof(vis));
	for(i=1;i<=n;i++)
	{
		dis[i]=map[x][i];
	}
	vis[x]=true;
	dis[x]=0;
	for(i=1;i<n;i++)
	{
		int p=x,min=INF;
		for(j=1;j<=n;j++)
		{
			if(!vis[j]&&dis[j]<min)
			{
				p=j;
				min=dis[j];
			}
		}
		vis[p]=true;
		for(j=1;j<=n;j++)
		{
			if(!vis[j]&&dis[j]>dis[p]+map[p][j])
			{
				dis[j]=dis[p]+map[p][j];
			}
		}
	}
	if(dis[n]==INF)
	{
		printf("Impossible\n");
	}
	else
	{
		printf("%d\n",dis[n]);
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	int cc=1;
	while(t--)
	{
		scanf("%d%d",&n,&m);
		init();
		int i,j;
		while(m--)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			if(map[u][v]>w)
			{
				map[u][v]=map[v][u]=w;
			}
		}
		printf("Case %d: ", cc++);
		f(1); 
	}
	return 0;
}