// #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()
// {
// int N, M;
// cin >> N >> M;
// for (int i = 0; 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;
// using T = tuple<ll, int, int>;
// priority_queue<T, vector<T>, greater<T>> pq;
// pq.push({0, 1, 0});
// while (!pq.empty())
// {
// ll cost;
// int node, c;
// tie(cost, node, c) = pq.top();
// pq.pop();
// }
// return 0;
// }
#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});
while (pq.size())
{
ll cost;
int node, c;
tie(cost, node, c) = pq.top();
pq.pop();
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;
if (case1 < dp[j.to])
{
dp[j.to] = case1;
pq.push({-dp[j.to], j.to, 0});
}
// Case 2: We flip j but not another edge of the same colour
ll case2 = j.p - cost;
if (case2 < dp[j.to])
{
dp[j.to] = case2;
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) || case3 < dp2[j.to][j.c])
{
dp2[j.to][j.c] = case3;
pq.push({-dp2[j.to][j.c], j.to, j.c});
}
}
}
}
}
cout << (dp[n] > INF ? -1 : dp[n]);
return 0;
}
