void solve()
{
    cout << setprecision(12);
    int n, m;
    cin >> n >> m;
    vpll cost(n, {INFF, INFF}); //cost will save minimum cost to reach ith node from node 0. cost[i].first is the
                                //cost and cost[i].second is  the number of edges used till now (basically k).

    cost[0] = {0, 0}; // cost to reach 0th node is 0 and 0 edges used till now.
    vector<vpll> graph(n); // graph saves cost-of-edge (w) and the connected vertex.
    rep(i, 0, m)
    {
        int x, y, w;
        cin >> x >> y >> w;
        x--;
        y--;
        graph[x].pb({w, y});
        graph[y].pb({w, x});
    }
    vb visited(n);
    int count = 0;
    while (count < n)
    {
        ll curr = INFF;
        int from = 0;
        rep(i, 0, n) // selecting the closest node. saved as 'from'.
        {
            if (cost[i].first < curr && !visited[i])
            {
                curr = cost[i].first;
                from = i;
            }
        }
        rep(i, 0, graph[from].size()) // calculating minimum cost for every node by using edges of current node ('from').
        {
            if (curr + graph[from][i].first * (cost[from].second + 1) < cost[graph[from][i].second].first)
            {
                cost[graph[from][i].second].first = curr + graph[from][i].first * (cost[from].second + 1);
                cost[graph[from][i].second].second = cost[from].second + 1;
            }
        }
        visited[from] = 1;
        count++;
    }
    rep(i, 0, n)
    {
        if (cost[i].first == INFF)
        {
            cout << -1 << "\n";
        }
        else
            cout << cost[i].first << "\n";
    }
    return;
}