#include<bits/stdc++.h>
using namespace std;
int M[1002][1002], Dist[1002], TF[1002];
void tree(int N)
{
int k, sum=0 ,i ,j;
for ( i = 2; i <= N; i ++)//把所有与顶点1相邻的边放入Dist数组
Dist[i] = M[1][i];
TF[1] = 1;
for ( i = 0; i < N-1; i ++)
{
int min1 = 9999, k = -1;
for ( j = 1; j <= N ; j ++)
{
if ( !TF[j] && Dist[j] < min1)//如果当前道路未被经过且权值小于当前最小的权值
{
min1 = Dist[j];//更改最小权值数值
k = j;//记录最小全职的下标
}
}
if ( k == -1 )//如果K的下标还是-1，说明没有通向该村庄的道路
{
cout<<"Impossible"<<endl;
return ;
}
TF[k] = 1;//记录该条道路已经过
sum += min1;
for ( j = 1; j <= N; j ++)
{
if ( !TF[j] && M[k][j] < Dist[j] )
Dist[j]=M[k][j];
}
}
cout<<sum<<endl;
}
int main()
{
int N, m, i, j;
while(cin>>N>>m)
{
for ( i = 1 ;i <= N ; i ++)
{
Dist[i] = 9999;
TF[i] = 0;//所有道路都标志为未经过
for ( j = 1; j <= N ; j ++)
M[i][j] = 9999;//所有道路权值初始化为无穷大
}
for ( i = 0 ; i < m ; i ++)//输入有权无向图的邻接矩阵
{
int a, b, c;
cin>>a>>b>>c;
M[a][b] = c;
M[b][a] = c;
}
tree(N);
}
return 0;
}