#include <iostream>
#include <vector>
#include <queue>

using namespace std;
const int INF = 1e9;

// A* 专用的状态结构体
struct State {
    int f, g, u;
    bool operator>(const State& other) const {
        return f > other.f; // 小顶堆，f 小的优先
    }
};

int main() {
    int n, m;
    cin >> n >> m;
    vector<vector<pair<int, int>>> adj(n + 1);
    vector<vector<pair<int, int>>> adjReverse(n + 1);
    for (int i = 0; i < m; ++i) {
        int a, b, t;
        cin >> a >> b >> t;
        adj[a].push_back({b, t});
        adjReverse[b].push_back({a, t});
    }
    int s, t, k;
    cin >> s >> t >> k;

    // 1. 反向 Dijkstra 计算 h(n)
    vector<int> h(n + 1, INF);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> jda;
    
    h[t] = 0;
    jda.push({0, t});
    while (!jda.empty()) {
        auto [d, u] = jda.top(); jda.pop();
        if (d > h[u]) continue;
        for (auto& edge : adjReverse[u]) {
            int v = edge.first, w = edge.second;
            if (h[u] + w < h[v]) {
                h[v] = h[u] + w;
                jda.push({h[v], v});
            }
        }
    }

    // 如果起点终点不连通，h[s] 依然是 INF
    if (h[s] == INF) {
        cout << -1 << endl;
        return 0;
    }

    // 2. 正向 A* 找第 K 短路
    if (s == t) k++; // 经典细节：起点终点重合，第一条路径长度为 0，需跳过

    priority_queue<State, vector<State>, greater<State>> astar;
    astar.push({h[s], 0, s}); // f = g + h = 0 + h[s]
    
    vector<int> count(n + 1, 0); // 记录每个点被弹出的次数

    while (!astar.empty()) {
        State curr = astar.top(); astar.pop();
        int u = curr.u;
        int g = curr.g;

        count[u]++;
        
        // 终点第 K 次出堆，就是答案！
        if (count[t] == k) {
            cout << g << endl;
            return 0;
        }

        // 如果某个点被弹出次数超过 k，说明再从它出发也找不到前 k 短了，优化剪枝
        if (count[u] > k) continue;

        for (auto& edge : adj[u]) {
            int v = edge.first, w = edge.second;
            // 压入新状态：g 增加实际边权，f 更新为新 g + h[v]
            astar.push({g + w + h[v], g + w, v});
        }
    }

    cout << -1 << endl;
    return 0;
}