#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<pair<ll, ll>> graph[100001];
ll du[100001], dv[100001], ds[100001], dp[2][100001], ans;
bool visited[100001];
void dijkstra1(ll start, ll arr[]) {
fill(visited, visited + 100001, false); //初始化visited为false
priority_queue<pair<ll, ll>> pq;
pq.push({0, start});
while (!pq.empty()) {
ll c, node;
tie(c, node) = pq.top(); //c: 从 start 到 node 的cost
pq.pop();
if (!visited[node]) {
arr[node] = -c; //距离
visited[node] = true;
for (auto& i : graph[node]) pq.push({c - i.second, i.first});
}
}
}
void dijkstra2(ll start, ll end) {
fill(dp[0], dp[0] + 100001, LLONG_MAX / 2); //初始化dp[0], minimum distance from u to any node x such that x is in a path from s to i in the DAG
fill(dp[1], dp[1] + 100001, LLONG_MAX / 2); //初始化dp[1], minimum distance from v to any node x such that x is in a path from s to i in the DAG
fill(visited, visited + 100001, false); //初始化visited为false
priority_queue<pair<ll, pair<ll, ll>>> pq;
pq.push({0, {start, 0}}); //cost, curr node, pre node
dp[0][0] = dp[1][0] = LLONG_MAX/ 2;
while (!pq.empty()) {
ll c, node, par;
pair<ll, ll> p;
tie(c, p) = pq.top();
tie(node, par) = p;
pq.pop();
if (!visited[node]) {
visited[node] = true;
ds[node] = -c;
dp[0][node] = min(du[node], dp[0][par]);
dp[1][node] = min(dv[node], dp[1][par]);
for (auto i : graph[node]) pq.push({c - i.second, {i.first, node}}); //
} else if (-c == ds[node]) { //另外一条最短路近
if (min(du[node], dp[0][par]) + min(dv[node], dp[1][par]) <= dp[0][node] + dp[1][node]) {
dp[0][node] = min(du[node], dp[0][par]);
dp[1][node] = min(dv[node], dp[1][par]);
}
}
}
ans = min(ans, dp[0][end] + dp[1][end]);
}
int main() {
iostream::sync_with_stdio(false);
cin.tie(0);
// freopen("CommuterPass.in", "r", stdin);
// freopen("CommuterPass.out", "w", stdout);
ll n, m, s, t, u, v;
cin >> n >> m >> s >> t >> u >> v;
for (int i = 0; i < m; i++) {
ll a, b, c;
cin >> a >> b >> c;
graph[a].push_back({b, c});
graph[b].push_back({a, c});
}
dijkstra1(u, du); //u出发，到所有其他点的距离，保存在du中。
dijkstra1(v, dv); //v出发，到所有其他点的距离，保存在dv中。
ans = du[v]; //ans是u到v的最短路径
dijkstra2(s, t);
dijkstra2(t, s);
cout << ans << '\n';
return 0;
}
