// https://oj.uz/problem/view/JOI20_ho_t4
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 210;
const int maxm = 5e4+10;
const int inf = 1e18;
const int comp = 1e17;
int N, M;
vector<pair<int, int>> g[maxn][2];
tuple<int, int, int, int> ar[maxm];
int dist[maxn];
bool proc[maxn];
int ans = inf;
int P[maxn][4];
void dijkstra(int A, bool b)
{
for(int i = 1; i <= N; i++)
{
dist[i] = 0;
proc[i] = 0;
}
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
pq.push({1, A});
dist[A] = 1;
while(!pq.empty())
{
int x = pq.top().second;
pq.pop();
if(proc[x]) continue;
proc[x] = 1;
for(auto[v, w] : g[x][b])
{
if(dist[v] == 0 || dist[x]+w < dist[v])
{
dist[v] = dist[x]+w;
pq.push({dist[v], v});
}
}
}
}
int32_t main()
{
cin >> N >> M;
for(int i = 1; i <= M; i++)
{
int u, v, c, d;
cin >> u >> v >> c >> d;
ar[i] = make_tuple(u, v, c, d);
g[u][0].push_back({v, c});
g[v][1].push_back({u, c});
}
if(M <= 1000)
{
for(int i = 0; i <= M; i++)
{
int D = 0;
if(i != 0)
{
auto[u, v, c, d] = ar[i];
for(auto it = g[u][0].begin(); it != g[u][0].end(); it++)
{
if(it->first == v && it->second == c)
{
g[u][0].erase(it);
break;
}
}
g[v][0].push_back({u, c});
D = d;
}
dijkstra(1, 0);
int p1 = dist[N]-1;
dijkstra(N, 0);
int p2 = dist[1]-1;
if(p1 != -1 && p2 != -1) ans = min(ans, p1+p2+D);
if(i != 0)
{
auto[u, v, c, d] = ar[i];
for(auto it = g[v][0].begin(); it != g[v][0].end(); it++)
{
if(it->first == u && it->second == c)
{
g[v][0].erase(it);
break;
}
}
g[u][0].push_back({v, c});
}
}
cout << (ans > comp ? -1 : ans);
return 0;
}
dijkstra(1, 0);
for(int i = 1; i <= N; i++)
{
P[i][0] = dist[i]-1;
if(P[i][0] == -1) P[i][0] = inf;
}
dijkstra(N, 1);
for(int i = 1; i <= N; i++)
{
P[i][1] = dist[i]-1;
if(P[i][1] == -1) P[i][1] = inf;
}
dijkstra(N, 0);
for(int i = 1; i <= N; i++)
{
P[i][2] = dist[i]-1;
if(P[i][2] == -1) P[i][2] = inf;
}
dijkstra(1, 1);
for(int i = 1; i <= N; i++)
{
P[i][3] = dist[i]-1;
if(P[i][3] == -1) P[i][3] = inf;
}
ans = P[N][0]+P[1][2];
for(int i = 1; i <= M; i++)
{
auto[u, v, c, d] = ar[i];
ans = min(ans, min(P[N][0], P[v][0]+ min(P[v][1], P[u][1]+c)) + min(P[1][2], P[v][2]+ min(P[v][3], P[u][3]+c)) +d);
}
cout << (ans > comp ? -1 : ans);
return 0;
}
// sub 2 -> ok after inverting you can still use the other way
// for each guy, have P(1, i) P(i, N) P(N, i) P(i, 1). For each V ans = +1, min(+2, P(U, N)+d+c), +3, min(+4, P(U,1)+d+c)
// sub 3 -> choose one and just get to the end
