#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 100002
#define N 1009
#define INF 1000000009
#define eps 0.00000001
/*
第K短路问题
1.将所有边反向建一个反图，然后求终点到各店的最短距离
2.优先队列 A*算法 每出队一个元素 cnt[cur]++
当cnr[destnation]==k 的时候 返回cur.g 就是第k短路的长度
*/
struct edge
{
	int to, cost, nxt;
}E[MAXN * 2];
struct node
{
	int g, h, pos;//g是当前点到源点的距离估值 h是当前点到终点的距离估值
	bool operator<(const node& rhs) const
	{
		return g + h < (rhs.g + rhs.h);
	}
};
int head1[N], head2[N];
int n, m, from, to, lowcost[N], cnt = 0;
void addedge(int u, int v, int w)
{
	E[cnt].to = v;
	E[cnt].cost = w;
	E[cnt].nxt = head1[u];
	head1[u] = cnt++;

	E[cnt].to = u;
	E[cnt].cost = w;
	E[cnt].nxt = head2[v];
	head2[v] = cnt++;
}
void Dijkstra()
{
	int vis[N];
	memset(vis, 0, sizeof(vis));
	memset(lowcost, INF, sizeof(lowcost));
	lowcost[to] = 0;
	for (int i = 0; i < n; i++)
	{
		int Minc = INF, k = -1;
		for (int j = 0; j < n; j++)
		{
			if (!vis[j] && lowcost[j] < Minc)
				Minc = lowcost[j], k = j;
		}
		vis[k] = true;
		for (int j = head2[k]; j != -1; j = E[j].nxt)
		{
			int v = E[j].to, w = E[j].cost;
			if (!vis[v] && lowcost[v] > lowcost[k] + w)
			{
				lowcost[v] = lowcost[k] + w;
			}
		}
	}
}
int Astar(int k)
{
	int num[N];
	memset(num, 0, sizeof(num));
	node tmp, f;
	priority_queue<node> q;
	tmp.g = 0;
	tmp.h = lowcost[from];
	tmp.pos = from;
	q.push(tmp);
	while (!q.empty())
	{
		f = q.top();
		q.pop();
		num[f.pos]++;
		if (num[f.pos] > k)
			continue;
		if (num[to] == k)
			return f.g;
		for (int i = head1[f.pos]; i != -1; i = E[i].nxt)
		{
			int v = E[i].to;
			tmp.pos = v;
			tmp.g = f.g + E[i].cost;
			tmp.h = lowcost[v];
			q.push(tmp);
		}
	}
	return -1;
}
int main()
{
	int a, b, c, k;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		memset(head1, -1, sizeof(head1));
		memset(head2, -1, sizeof(head2));
		cnt = 0;
		for (int i = 0; i < m; i++)
			scanf("%d%d%d", &a, &b, &c), addedge(a, b, c);
		scanf("%d%d%d", &from, &to, &k);
		if (from == to)
			k++;
		Dijkstra();
		printf("%d\n", Astar(k));
	}
	return 0;
}