#include<bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=2e5+10;
int fa[maxn],ran[maxn],n,m;     //并查集中的父结点数组和秩序数组
struct node
{
    int l,r,val;        //左右端点及权值
}arr[maxn];
void init()
{
    for(int i=1;i<=n;++i)
        fa[i]=i,ran[i]=0;
}
int ffind(int x)
{
    return x==fa[x]?x:fa[x]=ffind(fa[x]);
}
void unite(int x,int y)
{
    int fx=ffind(x);
    int fy=ffind(y);
    if(fx==fy)      return;
    if(ran[fx]<ran[fy]) fa[fx]=fy;
    else{
        fa[fy]=fx;
        if(fa[fx]==fa[fy])  ran[fx]++;
    }
}
bool operator < (node &a,node &b)
{
    return a.val<b.val;         //重载小于号，按权值从小到大排序
}
int Kruskal()
{
    int res=0,tot=0;
    for(int i=1;i<=m;++i)
    {
        if(ffind(arr[i].l)!=ffind(arr[i].r))
        {
            tot++;
            res+=arr[i].val;
            unite(arr[i].l,arr[i].r);
        }
        if(tot==n-1)    return res;
    }
    return inf;
}

int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        for(int i=1;i<=m;++i)
            cin>>arr[i].l>>arr[i].r>>arr[i].val;
        sort(arr+1,arr+m+1);
        init();
        int res=Kruskal();
        res==inf?cout<<"impossible"<<endl:cout<<res<<endl;
    }
}