#include <bits/stdc++.h>
using namespace std;
const long long oo = 1e18;
const int MAXN = 1e5 + 7;
vector<pair<int, int>> G[MAXN];
long long cost[MAXN][4];
long long pass[MAXN][2];
long long passprice;
int n;
void preprocces(int a, int x){
for(int i = 1; i <= n; i++) pass[i][x] = oo;
priority_queue<pair<long long, int>> Q;
Q.push({0, a});
while(Q.size()){
auto [d, v] = Q.top();
Q.pop();
if(pass[v][x] <= -d) continue;
pass[v][x] = -d;
for(auto [u, c] : G[v]){
if(pass[u][x] <= pass[v][x] + c) continue;
Q.push({-(pass[v][x] + c), u});
}
}
}
void dijkstra(int a){
// stan 0 jeszcze nie weszlismy na sciezke pass'a
// stan 1 weszlismy na sciezke pass'a ale idziemy w gore
// stan 2 weszlismy na sciezka pass'a ale idziemy w dol
// stan 3 wyszlismy z sciezki pass'a
priority_queue<tuple<long long, int, int>> Q;
Q.push({0, a, 0});
while(Q.size()){
auto [d, v, s] = Q.top(); Q.pop();
if(cost[v][s] <= -d) continue;
cost[v][s] = -d;
if(s == 0){
if(cost[v][1] > -d) Q.push({d, v, 1});
if(cost[v][2] > -d) Q.push({d, v, 2});
}
if(s == 1 || s == 2){
if(cost[v][3] > -d) Q.push({d, v, 3});
}
for(auto [u, c] : G[v]){
if(s == 0 || s == 3){
if(cost[u][s] <= cost[v][s] + c) continue;
Q.push({-(cost[v][s] + c), u, s});
}
else if(s == 1){
if(cost[u][s] <= cost[v][s]) continue;
if(pass[v][0] + c + pass[u][1] != passprice) continue;
//cout << pass[v][0] << " " << pass[u][1] << endl;
//cout << "s: " << s << " u: " << u << " v: " << v << " cost: " << cost[v][s] << endl;
if(pass[u][0] > pass[v][0])
Q.push({-cost[v][s], u, s});
}
else if(s == 2){
if(cost[u][s] <= cost[v][s]) continue;
if(pass[u][0] + c + pass[v][1] != passprice) continue;
//cout << "s: " << s << " u: " << u << " v: " << v << " cost: " << cost[v][s] << endl;
if(pass[u][1] > pass[v][1])
Q.push({-cost[v][s], u, s});
}
}
}
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
int m, s, t, u, v;
cin >> n >> m >> s >> t >> u >> v;
for(int i = 1; i <= m; i++){
int a, b, c;
cin >> a >> b >> c;
G[a].push_back({b, c});
G[b].push_back({a, c});
}
preprocces(s, 0);
preprocces(t, 1);
passprice = pass[t][0];
for(int i = 1; i <= n; i++)
cost[i][0] = cost[i][1] = cost[i][2] = cost[i][3] = oo;
dijkstra(u);
cout << min({cost[v][0], cost[v][1], cost[v][2], cost[v][3]}) << "\n";
//cout << cost[1][3] << endl;
}
