#include <bits/stdc++.h>
using namespace std;

using ll = long long;
static const ll INF = 1e18;

// 最小費用流
struct edge {
  int to;
  ll cap, cost;
  int rev;
};

vector<ll> h, d; // ポテンシャル, 最短距離
vector<int> prevv, preve; // 直前の頂点と辺

void add_edge(vector<vector<edge>> &G, int from, int to, ll cap, ll cost) {
  G[from].emplace_back((edge) {to, cap, cost, (int) G[to].size()});
  G[to].emplace_back((edge) {from, 0, -cost, (int) G[from].size() - 1});
}

// sからtへの流量fの最小費用流(流せない場合は-1)
ll min_cost_flow(vector<vector<edge>> &G, int s, int t, ll f) {
  ll res = 0;
  h.assign(G.size(), 0);
  prevv.assign(G.size(), 0);
  preve.assign(G.size(), 0);
  while (f > 0) {
    priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
    d.assign(G.size(), INF);
    d[s] = 0;
    pq.push(make_pair(0, s));
    while (!pq.empty()) {
      pair<ll, int> p = pq.top();
      pq.pop();
      int v = p.second;
      if (d[v] < p.first) continue;
      for (int i = 0; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && d[e.to] > d[v] + e.cost + h[v] - h[e.to]) {
          d[e.to] = d[v] + e.cost + h[v] - h[e.to];
          prevv[e.to] = v;
          preve[e.to] = i;
          pq.push(make_pair(d[e.to], e.to));
        }
      }
    }
    if (d[t] == INF) return -1;
    for (int i = 0; i < G.size(); i++) h[i] += d[i];
    
    // s-t間最短路に流す
    int nowf = f;
    for (int i = t; i != s; i = prevv[i]) {
      nowf = min<ll>(nowf, G[prevv[i]][preve[i]].cap);
    }
    f -= nowf;
    res += nowf * h[t];
    for (int i = t; i != s; i = prevv[i]) {
      edge &e = G[prevv[i]][preve[i]];
      e.cap -= nowf;
      G[i][e.rev].cap += nowf;
    }
  }
  return res;
}

int main() {
  ll n, m;
  cin >> n >> m;
  
  vector<vector<edge>> G(n);
  for (int i = 0; i < m; i++) {
    ll u, v, c, d;
    cin >> u >> v >> c >> d;
    u--;
    v--;
    add_edge(G, u, v, 1, c);
    add_edge(G, u, v, 1, d);
    add_edge(G, v, u, 1, c);
    add_edge(G, v, u, 1, d);
  }
  
  cout << min_cost_flow(G, 0, n - 1, 2) << '\n';
}