#include <bits/stdc++.h>
using namespace std;
const int N=100005;
const int MX=INT_MAX;
int n,m;
int a[5];
int d[6][N];
typedef pair<int,int> PII;
vector<PII> g[N];
void dij(int idx, int s) {
    priority_queue<PII, vector<PII>, greater<PII> > q;
    vector<bool> vis(n + 1, false);
    for (int i = 1; i <= n; ++i) d[idx][i] = INT_MAX;
    d[idx][s] = 0;
    q.push(make_pair(0, s));
    while (!q.empty()) {
        PII top = q.top(); q.pop();
        int u = top.second;
        if (vis[u]) continue;
        vis[u] = true;
        for (size_t i = 0; i < g[u].size(); ++i) {
            int v = g[u][i].first;
            int w = g[u][i].second;
            if (d[idx][v] > d[idx][u] + w) {
                d[idx][v] = d[idx][u] + w;
                q.push(make_pair(d[idx][v], v));
            }
        }
    }
}

int main(){
	cin>>n>>m;
	for(int i=1;i<=5;i++){
		cin>>a[i];
	}
	while(m--){
		int x,y,t;
		cin>>x>>y>>t;
		g[x].push_back({y, t});
		g[y].push_back({x, t});
	}
	for(int i=1;i<=5;i++){
		dij(i,a[i]);
	}
	int ans=0x3f3f3f3f;
	
    for (int city = 1; city <= n; ++city) {
        int sum = 0;
        for (int i = 1; i <= 5; ++i) {
            if (d[i][city] == MX) {
                sum = MX;
                break;
            }
            sum += d[i][city];
        }
        if (sum < ans) ans = sum;
    }

    cout << ans << "\n";
	return 0;
}
//n个点，m条边的无向无环联通图，有边权，求从1到abcde五个点的最短路（顺序任意）
//法一：贪心+dij（堆优化版）
//法二：Floyd：要爆空间 X
//法三：K短路：肯定有更简单的做法 X
//法四：暴搜：要爆时间 X