Bakery CodeForces - 707B（思维）

Bakery CodeForces - 707B

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.

Example

Input

5 4 2
1 2 5
1 2 3
2 3 4
1 4 10
1 5

Output

3

Input

3 1 1
1 2 3
3

Output

-1

Note

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

思路：这个题打眼一看好像是求什么最短路径，但是仔细一想会发现球什么最短路径，如果和面粉店有直接相连的不就是最短的了吗，这个题的题意和最短路径的题不同能一步解决的事，两步解决，肯定更长了，所以直接找直接相连的边就行，初始思路是用邻接表然后判断，但是一直wrong answ 13，然后看其他人这个点都错了很多次，后来才明白这个测试点就是如果两个点直接相连最短，但是这两点都是面粉店所以就不能选了，而用邻接表很难实现这个的判断，所以改用一种巧妙的标记

code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define INF 0x3f3f3f3f
//注意注意！！！如果两个店的起点和终点都是面粉店，即使最短但是不可以选
using namespace std;
int n,m,k;
int u[100100],v[100100],l[100100];//储存点，点，边长的信息
int vis[100100];//标记面粉店的数组
int main(){
    int i,j;
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&n,&m,&k);
    if(k==0){//没有面粉店直接输出-1返回
        printf("-1\n");
        return 0;
    }
    for(i = 0; i < m; i++){
        scanf("%d%d%d",&u[i],&v[i],&l[i]);//输入
    }
    for(i = 0; i < k; i++){
        int x;
        scanf("%d",&x);
        vis[x] = 1;//是面粉店的把这个点标记
    }
    int mind = INF;
    for(i = 0; i < m; i++){
        if(vis[u[i]]&&!vis[v[i]]||!vis[u[i]]&&vis[v[i]]){//判断每个直接的连边，如果一个时面粉店另一个不是，那么就可以
            mind = min(mind,l[i]);//这种写法巧妙的避免了使用邻接表出现两个点都是面粉店却无法判断的情况
        }
    }
    if(mind==INF)printf("-1\n");
    else printf("%d\n",mind);
    return 0;
}