#include<bits/stdc++.h>
#define int long long
#define pii pair<int, int>
#define fi first
#define se second
#define pri priority_queue
using namespace std;
const int maxn = 1e5;
const int inf = 1e15;
bool debug = 0;
// Problem B
// Subtask 1: If there are multiple optimal paths from S->T,
// Subtask 2: There is only one optimal path from S->T, just dijkstra and make the one optimmal path zeroes by keeping track the prev room
// Subtask 3: Possibly brute force
// Precompute all optimal paths from S->T, then for each path o(n) dijkstra o(nlogn) > o(n + n^2 logn)
int n, m, s, t, u, v;
int dist[maxn+5], pre[maxn+5];
map<pii, int> edge;
vector<int> adj[maxn+5];
int len(int a, int b) {
if (a > b) swap(a, b);
return edge[{a, b}];
}
void upd(int a, int b, int c) {
if (a > b) swap(a, b);
edge[{a, b}] = c;
}
signed main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin >> n >> m;
cin >> s >> t;
cin >> u >> v;
for (int i=1; i<=m; i++) {
int a, b, c;
cin >> a >> b >> c;
adj[a].push_back(b);
adj[b].push_back(a);
upd(a, b, c);
}
// subtask 2
pri<pii, vector<pii>, greater<pii>> pq;
for (int i=1; i<=n; i++) dist[i] = inf;
pq.push({0, s});
dist[s] = 0;
pre[s] = -1;
while (!pq.empty()) {
int now = pq.top().se, dis = pq.top().fi; pq.pop();
if (dist[now] < dis) continue;
for (int nxt: adj[now]) {
if (dis + len(now, nxt) >= dist[nxt]) continue;
dist[nxt] = dis + len(now, nxt);
pre[nxt] = now;
pq.push({dist[nxt], nxt});
}
}
if (debug) {
cout << "pre: " << endl;
for (int i=1; i<=n; i++) cout << pre[i] << " ";
cout << endl;
}
int cur = t;
while (pre[cur] != -1) {
upd(cur, pre[cur], 0);
cur = pre[cur];
}
for (int i=1; i<=n; i++) dist[i] = inf;
dist[u] = 0;
pq.push({0, u});
while (!pq.empty()) {
int now = pq.top().se, dis = pq.top().fi; pq.pop();
if (debug) cout << "now: " << now << " dis: " << dis << endl;
if (dist[now] < dis) continue;
for (int nxt: adj[now]) {
if (dis + len(now, nxt) >= dist[nxt]) continue;
dist[nxt] = dis + len(now, nxt);
pq.push({dist[nxt], nxt});
}
}
if (debug) {
cout << "dis : " << endl;
for (int i=1; i<=n; i++) cout << dist[i] << " ";
cout << endl;
}
cout << dist[v] << endl;
return 0;
}
