#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 100010, M = 200010, INF = 0x3f3f3f3f;

int n, m;
int p[N];

struct Edge
{
    int a, b, w;

    bool operator< (const Edge &W)const//重载比较符合表示以权值大小排序
    {
        return w < W.w;
    }
}edges[M];

int find(int x)
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int kruskal()
{
    sort(edges, edges+m);
    for(int i=1; i<=n; i++) p[i] = i;

    int res = 0, cnt = 0;
    for(int i=0; i<m; i++)
    {
        int a = edges[i].a, b = edges[i].b, w = edges[i].w;

        a=find(a), b=find(b);//让其等于各自祖宗结点
        if(a!=b)//判断两者是否连通，若不连通则
        {
            p[a] = b;//两个集合合并
            res += w;//res加的是，最小生成树边的权重之和
            cnt++;//当前加入多少边
        }
    }

    if(cnt < n-1) return INF;//判断一共加了多少条边，若是cnt小于n-1则说明不连通
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);

    for(int i=0; i<m; i++)
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }

    int t = kruskal();

    if(t == INF) puts("impossible");
    else printf("%d\n", t);

}