#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll INF = 1e18;
struct Edge {
int to, c;
ll p;
};
map<int, vector<Edge>> graph[100001];
ll dp[100001];
map<int, ll> dp2[100001], psum[100001];
int main() {
cin.tie(0)->sync_with_stdio(0);
int n, m;
cin >> n >> m;
for (int i = 1; i <= m; i++) {
int u, v, c;
ll p;
cin >> u >> v >> c >> p;
graph[u][c].push_back({v, c, p});
graph[v][c].push_back({u, c, p});
psum[u][c] += p;
psum[v][c] += p;
}
memset(dp, 0x3f, sizeof dp);
dp[1] = 0;
priority_queue<tuple<ll, int, int>> pq;
pq.push({0, 1, 0});
map<int, int> vis[n+1];
while (pq.size()) {
ll cost;
int node, c;
tie(cost, node, c) = pq.top();
pq.pop();
if (vis[node][c]) continue;
vis[node][c] = true;
if (c) {
if (dp2[node][c] != -cost) continue;
for (Edge i : graph[node][c]) {
// We can't flip i in this case
ll case1 = psum[node][c] - i.p;
if (case1 - cost < dp[i.to]) {
dp[i.to] = case1 - cost;
pq.push({-dp[i.to], i.to, 0});
}
}
} else {
if (dp[node] != -cost) continue;
for (auto& i : graph[node]) {
for (Edge j : i.second) {
// Case 1: We don't flip j
ll case1 = psum[node][j.c] - j.p - cost;
// Case 2: We flip j but not another edge of the same colour
ll case2 = j.p - cost;
// taking min of both cases
ll mnCase = min(case1, case2);
dp[j.to] = min(mnCase, dp[j.to]);
pq.push({-dp[j.to], j.to, 0});
// Case 3: We flip j and another edge of the same colour
ll case3 = -cost;
if (!dp2[j.to].count(j.c)) {
dp2[j.to][j.c] = case3;
pq.push({-dp2[j.to][j.c], j.to, j.c});
}
else {
dp2[j.to][j.c] = min(case3, dp2[j.to][j.c]);
pq.push({-dp2[j.to][j.c], j.to, j.c});
}
}
}
}
}
cout << (dp[n] > INF ? -1 : dp[n]);
return 0;
}
