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最新推荐文章于 2019-03-02 23:52:00 发布

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最新推荐文章于 2019-03-02 23:52:00 发布

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Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities.

To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are only k storages, located in different cities numbered a₁, a₂, ..., a_k.

Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another n - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.

Formally, Masha will pay x roubles, if she will open the bakery in some city b (a_i ≠ b for every 1 ≤ i ≤ k) and choose a storage in some city s (s = a_j for some 1 ≤ j ≤ k) and b and s are connected by some path of roads of summary length x (if there are more than one path, Masha is able to choose which of them should be used).

Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of k storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.

Input

The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 10⁵, 0 ≤ k ≤ n) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.

Then m lines follow. Each of them contains three integers u, v and l (1 ≤ u, v ≤ n, 1 ≤ l ≤ 10⁹, u ≠ v) meaning that there is a road between cities u and v of length of l kilometers .

If k > 0, then the last line of the input contains k distinct integers a₁, a₂, ..., a_k (1 ≤ a_i ≤ n) — the number of cities having flour storage located in. If k = 0 then this line is not presented in the input.

Output

Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.

If the bakery can not be opened (while satisfying conditions) in any of the n cities, print - 1 in the only line.

Examples

Input

Output

Input

3 1 1
1 2 3
3

Output

-1

Note

$\text{[math]}$

Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.

题意：有n座城市，给定m组城市之间的距离；又指明k座城市为仓库，小明在这n座城市中开一家面包店，需要一个仓库，但又不能开在仓库所在的城市，所以现在需要求k个仓库到某一座城市的最短距离！

思路：用一个标记数组记录仓库城市。然后遍历m组，判断连通的城市是否为一个无仓库城市，一个有仓库城市，且寻找最小路径即可！

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int u[100005],v[100005],l[100005],visit[100005];
int main()
{
    int i,n,m,k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        memset(visit,0,sizeof(visit));
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u[i],&v[i],&l[i]);
        }
        int point;
        for(i=0;i<k;i++)
        {
            scanf("%d",&point);
            visit[point] = 1;
        }
        int min = 2000000000;
        for(i=0;i<m;i++)
        {
            if(visit[u[i]] != visit[v[i]] && min > l[i])//1为仓里，0为面包店，不相等即符合，且寻找最小值
            {
                min = l[i];
            }
        }
        if(min == 2000000000)
            min = -1;
        printf("%d\n",min);
    }
    return 0;
}