/*
Kruskal 算法思路
1. 把所有边按权值从小到大排序O(mlogm)
2. for 所有边 a, b, w  O(m)
    if a, b不连通
        将这条边加入集合中

*/
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 200010;

int n, m;
int p[N];

struct Edge
{
    int a, b, w;

    bool operator < (const Edge& t) const
    {
        return w < t.w;
    }
}edges[N];

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

int main()
{
    cin >> 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};//下面思考一下重边和自环
    }

    sort(edges, edges + m);//按权值从小到大排序，重边这里加上判断a!=b过滤掉
    for (int i = 1; i <= n; i ++ ) p[i] = i;

    int res = 0, cnt = 0;//res存储所有边的权值之和，cnt存的是边数
    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);//如果是自环或回环路的话，a!=b过滤掉
        if (a != b)//a, b不连通
        {
            p[a] = b;//连通
            res += w;
            cnt ++ ;
        }
    }
    //重边，自环, 回环路都过滤掉了，边数就应该是n - 1 条,没有这么多条，正面最小生成树不存在,写小于< 是代码少点
    if (cnt < n - 1) puts("impossible");
    else printf("%d\n", res);

    return 0;
}