// P1629 / 教材例 10-3：有向正权图，题目保证任意两点互相可达。
// 输入 n m，随后 m 行 u v w；顶点从 1 编号。
#include <iostream>
#include <vector>
#include <queue>
#include <utility>
#include <functional>
using namespace std;
using ll = long long;
using State = pair<ll, int>;
struct Edge { int to; ll w; };
vector<ll> dijkstra(const vector<vector<Edge>> &g, int s) {
    const ll INF = (1LL << 60);
    vector<ll> dis(g.size(), INF);
    priority_queue<State, vector<State>, greater<State>> q;
    dis[s] = 0; q.push({0, s});
    while (!q.empty()) {
        State cur = q.top(); q.pop();
        ll d = cur.first; int u = cur.second;
        if (d != dis[u]) continue;
        for (const auto &e : g[u]) {
            ll nd = d + e.w;
            if (nd < dis[e.to]) {
                dis[e.to] = nd;
                q.push({nd, e.to});
            }
        }
    }
    return dis;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<Edge>> g(n + 1), rg(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v; ll w; cin >> u >> v >> w;
        g[u].push_back({v, w});
        rg[v].push_back({u, w});
    }
    vector<ll> out = dijkstra(g, 1);
    vector<ll> back = dijkstra(rg, 1);
    ll ans = 0;
    for (int i = 2; i <= n; ++i) ans += out[i] + back[i];
    cout << ans << '\n';
}