// P6464 传送门。原图 Floyd 一次，独立枚举一对零边。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n, m; if (!(cin >> n >> m)) return 0;
    const ll INF = 1LL << 60;
    vector<vector<ll>> d(n, vector<ll>(n, INF));
    for (int i = 0; i < n; ++i) d[i][i] = 0;
    for (int e = 0, u, v; e < m; ++e) {
        ll w; cin >> u >> v >> w; --u; --v;
        d[u][v] = d[v][u] = min(d[u][v], w);
    }
    for (int k = 0; k < n; ++k)
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                if (d[i][k] != INF && d[k][j] != INF)
                    d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
    ll ans = 0;
    for (int i = 0; i < n; ++i) for (int j = i+1; j < n; ++j) ans += d[i][j];
    for (int u = 0; u < n; ++u) for (int v = u+1; v < n; ++v) {
        ll sum = 0;
        for (int i = 0; i < n; ++i) for (int j = i+1; j < n; ++j)
            sum += min({d[i][j], d[i][u]+d[v][j], d[i][v]+d[u][j]});
        ans = min(ans, sum);
    }
    cout << ans << '\n';
}