// P4568 / 教材例 10-2：隐式分层图，最多 k 次免费。
// n m k；s t；m 行 u v w。城市从 0 编号，无向非负边。
// 课堂扩展约定：如果没有路线，输出 -1；原题未指定此情况的输出。
#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
#include <functional>
#include <algorithm>
using namespace std;
using ll = long long;
using State = tuple<ll, int, int>;
struct Edge { int to; ll w; };
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n, m, k, s, t;
    if (!(cin >> n >> m >> k >> s >> t)) return 0;
    vector<vector<Edge>> g(n);
    for (int i = 0; i < m; ++i) {
        int u, v; ll w; cin >> u >> v >> w;
        g[u].push_back({v, w});
        g[v].push_back({u, w});
    }
    const ll INF = (1LL << 60);
    vector<vector<ll>> dis(k + 1, vector<ll>(n, INF));
    priority_queue<State, vector<State>, greater<State>> q;
    dis[0][s] = 0;
    q.push({0, s, 0});
    while (!q.empty()) {
        State cur = q.top(); q.pop();
        ll d = get<0>(cur);
        int u = get<1>(cur), j = get<2>(cur);
        if (d != dis[j][u]) continue;
        for (const auto &e : g[u]) {
            int v = e.to;
            ll nd = d + e.w;
            if (nd < dis[j][v]) {
                dis[j][v] = nd;
                q.push({nd, v, j});
            }
            if (j < k && d < dis[j + 1][v]) {
                dis[j + 1][v] = d;
                q.push({d, v, j + 1});
            }
        }
    }
    ll ans = INF;
    for (int j = 0; j <= k; ++j) ans = min(ans, dis[j][t]);
    cout << (ans == INF ? -1 : ans) << '\n';
}