import java.io.*;
import java.util.*;
public class commuter_pass {
static int n, m, s, t, u, v;
static ArrayList<Edge>[] arr;
static long[] fromU, fromV;
public static void main(String[] args) throws IOException{
BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
//BufferedReader file = new BufferedReader(new FileReader("file.in"));
StringTokenizer st = new StringTokenizer(file.readLine());
n = Integer.parseInt(st.nextToken());
m = Integer.parseInt(st.nextToken());
st = new StringTokenizer(file.readLine());
s = Integer.parseInt(st.nextToken())-1;
t = Integer.parseInt(st.nextToken())-1;
st = new StringTokenizer(file.readLine());
u = Integer.parseInt(st.nextToken())-1;
v = Integer.parseInt(st.nextToken())-1;
arr = new ArrayList[n];
for (int i=0; i<n; i++){
arr[i] = new ArrayList<>();
}
for (int i=0; i<m; i++){
st = new StringTokenizer(file.readLine());
int s = Integer.parseInt(st.nextToken())-1;
int e = Integer.parseInt(st.nextToken())-1;
int c = Integer.parseInt(st.nextToken());
arr[s].add(new Edge(e, c));
arr[e].add(new Edge(s, c));
}
fromU = new long[n];
fromV = new long[n];
// First get distances from u and v to all other nodes
dij(u, fromU);
dij(v, fromV);
/*for (long i : fromV){
System.out.println(i);
}*/
long a = dij2(s, t);
/* Notice if we take all edges part of a SP from s to all other nodes, and we make them a directed edge, then we form a DAG. The definition of a SP is that there is no cycles, because if there exists a cycle, then it is NOT the SP.
Then, the tricky part is how to "find the best result over all possible SP". This is where DP on a DAG comes in.
Define: dpV[i] = the minimum cost for V to reach any node on the SP from start to i
dpU[i] = the minimum cost for U to reach any node on the SP from start to i
Note: dpV[i] and dpU[i] must be referencing the same SP. For example, If we have 2 equal paths 1 and 2 from s to x, node V to path 1 cost 1 and to path 2 cost 2, node U to path 1 cost 5 to path 2 cost 1. We cannot say the dpU[x] = 1 AND dpV[x] = 1, because node U achieves cost 1 with path 1, while node V achieves 1 with path 2. We cannot select both paths. The purpose of the DP is to find the best SP to select such that the value of dpV[end] + dpU[end] is minimial. In the previous example, the best result is dpU[x] = 1 and dpV[x] = 2, such that they both use path 2 and achieve a cost of 3 in total.
From the observation above, the transition of the DP is:
From the current node, the min cost to reach node V is min(dpV[parent], to go V directly from curr), same for node U. By looking at the parent, we know that the current node is part of (or the "next node") of the path that dpV[parent] and dpV[parent] is referencing. Thus, to transition, we can either take the best result from dpV[parent] or dpU[parent] (since they are part of the SP path from the start to current node), or go directly to U or V from the current node.
if (dpV[i] + dpU[i] < dpV[c])
dpV[i] = Math.min(dpV[i], Math.min(dpV[parent], fromV[])
*/
long b = dij2(t, s);
long ans = Math.min(fromU[v], Math.min(a, b));
System.out.println(ans);
}
public static void dij(int start, long[] dist){
PriorityQueue<State> pq = new PriorityQueue<>();
Arrays.fill(dist, (long)1e18);
pq.add(new State(start, -1, 0));
dist[start] = 0;
while (!pq.isEmpty()){
State curr = pq.poll();
if (dist[curr.id] != curr.cost){
continue;
}
for (Edge i : arr[curr.id]){
if (curr.cost + i.cost < dist[i.id]){
dist[i.id] = curr.cost + i.cost;
pq.add(new State(i.id, curr.id, curr.cost + i.cost));
}
}
}
}
public static long dij2(int start, int end){
long[] dist = new long[n];
long[] dpU = new long[n];
long[] dpV = new long[n];
PriorityQueue<State> pq = new PriorityQueue<>();
Arrays.fill(dist, (long)1e18);
Arrays.fill(dpU, (long)1e18);
Arrays.fill(dpV, (long)1e18);
boolean[] used = new boolean[n];
pq.add(new State(start, start, 0));
dist[start] = 0;
dpV[start] = fromV[start];
dpU[start] = fromU[start];
while (!pq.isEmpty()){
State curr = pq.poll();
if (dist[curr.id] != curr.cost){
continue;
}
if (!used[curr.id]){
used[curr.id] = true;
for (Edge i : arr[curr.id]){
if (curr.cost + i.cost <= dist[i.id]){
dist[i.id] = curr.cost + i.cost;
pq.add(new State(i.id, curr.id, curr.cost + i.cost));
}
}
}
if (dist[curr.id] == curr.cost){ // Is on SP
// 2 options, take DP value from parent, or go directly to u or v from curr node.
if (dpV[curr.id] + dpU[curr.id] > Math.min(fromV[curr.id], dpV[curr.par]) + Math.min(fromU[curr.id], dpU[curr.par])){
dpV[curr.id] = Math.min(fromV[curr.id], dpV[curr.par]);
dpU[curr.id] = Math.min(fromU[curr.id], dpU[curr.par]);
}
}
}
return dpV[end] + dpU[end];
}
static class State implements Comparable<State>{
int id, par;
long cost;
public State(int id, int par, long cost){
this.id = id;
this.par = par;
this.cost = cost;
}
public int compareTo(State o){
return Long.compare(this.cost, o.cost);
}
}
static class Edge{
int id;
long cost;
public Edge(int id, long cost){
this.id = id;
this.cost = cost;
}
}
}
