Codeforces Round #368 (Div. 2) B. Bakery

B. Bakery

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Masha wants to open her own bakery and bake muffins in one of then 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 onlyk storages, located in different cities numbereda₁, 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 anothern - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay1 ruble.

Formally, Masha will pay x roubles, if she will open the bakery in some cityb (a_i ≠ b for every1 ≤ i ≤ k) and choose a storage in some citys (s = a_j for some1 ≤ j ≤ k) and b and s are connected by some path of roads of summary lengthx (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 ofk 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 integersn, m andk (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 integersu, v andl (1 ≤ u, v ≤ n,1 ≤ l ≤ 10⁹,u ≠ v) meaning that there is a road between citiesu and v of length ofl kilometers .

If k > 0, then the last line of the input containsk distinct integers a₁, a₂, ..., a_k (1 ≤ a_i ≤ n) — the number of cities having flour storage located in. If k = 0 then this lineis 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 then cities, print  - 1 in the only line.

Examples

Input

5 4 21 2 51 2 32 3 41 4 101 5

Output

3

Input

3 1 11 2 33

Output

-1

Note

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

题解：题目大意为Masha 想在N个城市中开个面包店，其中这N城市中有K个城市有面粉供应，但该国有法律明文规定在有面粉供应的城市中禁止开店（至于为什么有这么坑爹的规定，这个就不知道了……），面粉供应点与面包店距离要尽量小，这样可以减少运费，毕竟Masha可是个节俭的人…
       因此该题可以看做在N个点中有K个特殊点，从这K个特殊点中找    到普通点的最小距离……

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<stdlib.h>#include<time.h>#include<math.h>#include<queue>;#include<stack>;#define INF 0x3f3f3f3fusing namespace std;int x[100005],y[100005],z[100005],v[100005];           //x,y,z,分别存储起点 终点  和距离  V做标记，表示是否为特殊点int main(){    int n,m,k,ans=INF,a;    scanf("%d%d%d",&n,&m,&k);    memset(v,0,sizeof(v));    for(int i=1;i<=m;i++)scanf("%d%d%d",&x[i],&y[i],&z[i]);          // 输入……    for(int i=1;i<=k;i++){scanf("%d",&a);v[a]=1;}               //  标记……    for(int i=1;i<=m;i++)                               //   找最小距离    {        if(v[x[i]]^v[y[i]])                              //   按位异或  X与Y中一个为特殊点一个为普通点的情况            ans=min(ans,z[i]);    }    if(ans==INF)        printf("-1\n");    else        printf("%d\n",ans);}