#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MXN = 1e5 + 1, MXM = 2e5 + 1;
const ll INF = 1e18 + 5;
struct state {
int node, colour;
ll cur_dist;
state(int _node = 0, int _colour = 0, ll _cur_dist = 0) : node(_node), colour(_colour), cur_dist(_cur_dist) {}
bool operator<(const state &other) const {
return cur_dist > other.cur_dist;
}
};
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int N, M;
cin >> N >> M;
map<int, ll> sum[N];
vector<array<int, 3>> g[N];
map<int, vector<pair<int, int>>> edges[N];
for (int i = 0; i < M; i++) {
int a, b, c, p;
cin >> a >> b >> c >> p;
a--, b--;
// {v, colour of edge, cost of deleting edge}
g[a].emplace_back(array<int, 3>{b, c, p});
g[b].emplace_back(array<int, 3>{a, c, p});
// Sum of costs of all edges incident to a node with a particular colour
sum[a][c] += p;
sum[b][c] += p;
// All edges incident to a node with a particular colour
edges[a][c].emplace_back(b, p);
edges[b][c].emplace_back(a, p);
}
vector<ll> dist(N, INF);
dist[0] = 0;
map<int, ll> dist_col[N];
priority_queue<state> pq;
pq.emplace(0, -1, 0); // -1 represents free will, no colour restriction
while (!pq.empty()) {
auto [u, colour, cur_dist] = pq.top();
pq.pop();
if (colour == -1) {
if (dist[u] != cur_dist) continue;
for (const auto &[v, edge_col, edge_cost] : g[u]) {
// 2.1 and 2.2
ll cost1 = min(cur_dist + sum[u][edge_col] - edge_cost, cur_dist + edge_cost);
if (cost1 < dist[v]) {
dist[v] = cost1;
pq.emplace(v, -1, dist[v]);
}
// 2.3
if (!dist_col[v].count(edge_col) || cur_dist < dist_col[v][edge_col]) {
dist_col[v][edge_col] = cur_dist;
pq.emplace(v, edge_col, cur_dist);
}
}
} else {
if (cur_dist != dist_col[u][colour]) continue;
for (const auto &[v, edge_cost] : edges[u][colour]) {
ll cost = cur_dist + sum[u][colour] - edge_cost;
if (cost < dist[v]) {
dist[v] = cost;
pq.emplace(v, -1, cost);
}
}
}
}
cout << (dist[N - 1] < INF ? dist[N - 1] : -1);
}
/*
Observations:
1. Always possible to change the colour of an edge so that it is different from all other edges connecting u if the answer is not -1 (?)
2. Three cases if current edge E(u, v) is of colour C:
2.1: We use E by changing colours of all other edges with C to go to v. At v, it's okay to use an edge of colour C since it gives us a worse cost than 2.2.
2.2: We change E's colour and go to v. Don't change the colour of any other edge incident to u with colour C. We can use an edge of any colour from v.
This gives us a worse cost if we use an edge of colour C form v because we double count the cost for this edge. Case 2.3 handles this.
2.3: We change E's colour and use an edge of colour C from v.
*/
