#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define vi vector<long long>
#define pr pair<ll, ll>
const ll INF = 2e15; // Set safely above K (10^15) to prevent overflow during addition
void dij(int n, int start, vi &dist, const vector<vector<pr>>& adj) {
dist.assign(n + 1, INF);
priority_queue<pr, vector<pr>, greater<pr>> pq;
dist[start] = 0;
pq.push({0, start});
while (!pq.empty()) {
auto [d, u] = pq.top(); pq.pop();
if (d > dist[u]) continue;
for (auto& edge : adj[u]) {
if (dist[u] + edge.second < dist[edge.first]) {
dist[edge.first] = dist[u] + edge.second;
pq.push({dist[edge.first], edge.first});
}
}
}
}
void solution() {
int n, m;
if (!(cin >> n >> m)) return;
ll S, T, L, K;
cin >> S >> T >> L >> K;
vector<vector<pr>> adj(n + 1);
for (int i = 0; i < m; i++) {
int u, v;
ll c;
cin >> u >> v >> c;
adj[u].push_back({v, c});
adj[v].push_back({u, c});
}
vi dS, dT;
dij(n, S, dS, adj);
dij(n, T, dT, adj);
// Case: King is already happy with existing lines
if (dS[T] <= K) {
cout << (ll)n * (n - 1) / 2 << endl; // All possible ways [cite: 13]
return;
}
ll ans = 0;
// Standard O(N^2) loop to check all pairs {u, v} [cite: 13]
for (int u = 1; u <= n; u++) {
for (int v = u + 1; v <= n; v++) {
// New path can go S -> u -> v -> T OR S -> v -> u -> T
ll pathA = dS[u] + L + dT[v];
ll pathB = dS[v] + L + dT[u];
if (pathA <= K || pathB <= K) {
ans++;
}
}
}
cout << ans << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
solution();
return 0;
}
