/*
If you're looking at my code for solution,
your brute-force code with a bit optimization is the solution.
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pl;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tt;
#define all(a) a.begin(), a.end()
#define filter(a) a.erase(unique(all(a)), a.end())
const int E = 5e4 + 4;
const int mn = 202;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
int direct[E], a[E], b[E], cost[E], revCost[E], n;
ll distanc[mn][mn], dist[mn];
vector<tt> adj[mn];
bool vis[mn];
ll dijkstra (int s, int t) {
for (int i = 1; i <= n; i++) dist[i] = LLONG_MAX, vis[i] = 0;
dist[s] = 0;
priority_queue<pl> pq; pq.emplace(0, s);
while (pq.size()) {
int u = pq.top().second; pq.pop();
if (vis[u]) continue;
if (u == t) return dist[t];
vis[u] = 1;
for (tt it : adj[u]) {
int v, id, way; tie(v, id, way) = it;
if (way != direct[id]) continue;
if (dist[u] + cost[id] < dist[v]) {
dist[v] = dist[u] + cost[id];
pq.emplace(-dist[v], v);
}
}
}
return LLONG_MAX;
}
ll add (ll a, ll b) {
return (max(a, b) == LLONG_MAX ? LLONG_MAX : a + b);
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int m; cin >> n >> m;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
distanc[i][j] = (i == j ? 0 : LLONG_MAX);
for (int i = 0; i < m; i++) {
cin >> a[i] >> b[i] >> cost[i] >> revCost[i];
adj[a[i]].emplace_back(b[i], i, 0);
adj[b[i]].emplace_back(a[i], i, 1);
distanc[a[i]][b[i]] = min(distanc[a[i]][b[i]], (ll)cost[i]);
}
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
distanc[i][j] = min(distanc[i][j], add(distanc[i][k], distanc[k][j]));
vector<int> cand(m);
for (int i = 0; i < m; i++) cand[i] = i;
shuffle(all(cand), rng);
ll ans = add(distanc[1][n], distanc[n][1]);
for (int i : cand) {
ll tryOTN = min(distanc[1][n], add(add(distanc[1][b[i]], cost[i]), distanc[a[i]][n]));
ll tryNTO = min(distanc[n][1], add(add(distanc[n][b[i]], cost[i]), distanc[a[i]][1]));
if (add(add(tryOTN, tryNTO), revCost[i]) < ans) { // only optimize when possible
direct[i] = 1;
ll dist = add(dijkstra(1, n), dijkstra(n, 1));
ans = min(ans, add(dist, revCost[i]));
direct[i] = 0;
}
}
cout << (ans == LLONG_MAX ? -1 : ans);
return 0;
}
