/*
* Key Observation: 只能顺着S->T走一段免费的或者逆着S->T走一段免费的; 不可能顺着+逆着，否则这一定不是S->T的最短路的一部分
*/
#include <bits/stdc++.h>
#define int long long
#define INF 1e16
#define pi pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
const int maxN = 1e5+5;
struct cmp{
bool operator() (pi a, pi b){
return a.fi > b.fi;
}
};
int N,M,S,T,U,V,res;
vector<pi> adj[maxN];
int distS[maxN],distS2[maxN],distT[maxN],distT2[maxN],distU[maxN],distV[maxN],dpU[maxN],dpV[maxN];
void dijk(int dist[], int src){
for(int i = 1; i <= N; ++i)dist[i] = INF;
bool vis[maxN];
memset(vis,false,sizeof(vis));
priority_queue<pi,vector<pi>,cmp> Q;
Q.push(pi(0,src));
dist[src] = 0;
while(!Q.empty()){
pi p = Q.top(); Q.pop();
int d = p.fi; int loc = p.se;
if(vis[loc])continue;
vis[loc] = true;
for(auto x: adj[loc]){
if(dist[loc] + x.se < dist[x.fi]){
dist[x.fi] = dist[loc] + x.se;
if(!vis[x.fi]){
Q.push(pi(dist[x.fi],x.fi));
}
}
}
}
}
void dijk2(int dist[], int src, int end){
for(int i = 1; i <= N; ++i)dist[i] = INF;
bool vis[maxN];
memset(vis,false,sizeof(vis));
for(int i = 1; i <= N; ++i)dpU[i] = dpV[i] = INF;
priority_queue<pi,vector<pi>,cmp> Q;
Q.push(pi(0,src));
dist[src] = 0;
dpU[src] = distU[src];
dpV[src] = distV[src];
while(!Q.empty()){
pi p = Q.top(); Q.pop();
int d = p.fi; int loc = p.se;
if(vis[loc])continue;
vis[loc] = true;
for(auto x: adj[loc]){
if(dist[loc] + x.se < dist[x.fi]){
dist[x.fi] = dist[loc] + x.se;
if(!vis[x.fi]){
Q.push(pi(dist[x.fi],x.fi));
}
}
if(distS[x.fi] + distT[x.fi] == distS[T]){
//There must be a shortest path from S to T that passes thru x.fi if the above is satisfied
dpU[x.fi] = min(dpU[x.fi],min(dpU[loc],distU[x.fi]));
dpV[x.fi] = min(dpV[x.fi],min(dpV[loc],distV[x.fi]));
}
}
}
res = min(res,dpU[end]+dpV[end]);
// for(int i = 1; i <= N; ++i)res = min(res,dpU[i]+dpV[i]);
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
//freopen("../test.in","r",stdin);
cin >> N >> M >> S >> T >> U >> V;
for(int i = 1; i <= M; ++i){
int a,b,c; cin >> a >> b >> c;
adj[a].pb(pi(b,c));
adj[b].pb(pi(a,c));
}
dijk(distU,U);
dijk(distV,V);
dijk(distS,S);
dijk(distT,T);
res = distU[V]; //Min dist without commuter pass
dijk2(distS2,S,T);
dijk2(distT2,T,S);
cout << res << '\n';
}
